JK-TK/Newdata huggingface dataset:
Calculate the atomic mass of ‘M’ Equation Chemical equation NaOH(aq) + HCl(aq) -> KCl(aq) + H2O(l) Moles of NaOH = Molarity x volume=> 0.1 x20 = 0.002 moles 1000 1000 Mole ratio HCl; NaOH = 1:1 Excess moles of HCl = 0.002 moles 25cm3 -> 0.002 moles 1000cm3 -> 1000 x 0.002 = 0.08moles 25cm3 Original moles of HCl = Molarity x volume => 1M x 1litre = 1.0 moles Moles of HCl reacted with MCO3 = 1.0 - 0.08 moles = 0.92moles Chemical equation MCO3(s) + 2HCl(aq) -> MCl2 (aq) + CO2 (g) + H2O(l) Mole ratio MCO3(s) : HCl(aq) =1:2 Moles of MCO3 = 0.92moles => 0.46moles 2 Molar mass of MCO3= mass => 46g = 100 g moles 0.46moles M= MCO3 - CO3 =>100g – (12+ 16 x3 = 60) = 40
52 52 6. 25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre.
25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre. Mole ratio Fe2+ :Mn04- = 5:1 Moles Mn04- used = 0.02 x 15 = 3.0 x 10-4 moles 1000 Moles Fe2+ = 3.0 x 10-4 moles = 6.0 x 10-5 moles 5 Molarity of Fe2+ = 6.0 x 10-4 moles x 1000 = 2.4 x 10-3 moles l-1 25 Since Zinc reduces Fe3+ to Fe2+ in the mixture: Moles Mn04- that reacted with all Fe2+= 0.02 x 19 = 3.8 x 10-4 moles 1000 Moles of all Fe2+ = 3.8 x 10-4 moles = 7.6 x 10-5 moles 5 Moles of Fe3+ = 3.8 x 10-4 - 6.0 x 10-5 = 1.6 x 10-5 moles Molarity of Fe3+ = 1.6 x 10-5 moles x 1000 = 4.0 x 10-4 moles l-1 25
14.0.0 ORGANIC CHEMISTRY I (HYDROCARBONS) (25 LESSONS) Introduction to Organic chemistry Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made. Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent.
Mole ratio Fe2+ :Mn04- = 5:1 Moles Mn04- used = 0.02 x 15 = 3.0 x 10-4 moles 1000 Moles Fe2+ = 3.0 x 10-4 moles = 6.0 x 10-5 moles 5 Molarity of Fe2+ = 6.0 x 10-4 moles x 1000 = 2.4 x 10-3 moles l-1 25 Since Zinc reduces Fe3+ to Fe2+ in the mixture: Moles Mn04- that reacted with all Fe2+= 0.02 x 19 = 3.8 x 10-4 moles 1000 Moles of all Fe2+ = 3.8 x 10-4 moles = 7.6 x 10-5 moles 5 Moles of Fe3+ = 3.8 x 10-4 - 6.0 x 10-5 = 1.6 x 10-5 moles Molarity of Fe3+ = 1.6 x 10-5 moles x 1000 = 4.0 x 10-4 moles l-1 25
14.0.0 ORGANIC CHEMISTRY I (HYDROCARBONS) (25 LESSONS) Introduction to Organic chemistry Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made. Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent. It is able to form stable covalent bonds with itself and many non-metals like hydrogen, nitrogen ,oxygen and halogens to form a variety of compounds. This is because: (i) carbon uses all the four valence electrons to form four strong covalent bond. (ii)carbon can covalently bond to form a single, double or triple covalent bond with itself. (iii)carbon atoms can covalently bond to form a very long chain or ring. When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only.
(ii)carbon can covalently bond to form a single, double or triple covalent bond with itself. (iii)carbon atoms can covalently bond to form a very long chain or ring. When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only. Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as: (i) Alkanes (ii) Alkenes (iii) Alkynes (i) Alkanes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule. The carbon atoms are linked by single bond to each other and to hydrogen atoms.
When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only. Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as: (i) Alkanes (ii) Alkenes (iii) Alkynes (i) Alkanes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule. The carbon atoms are linked by single bond to each other and to hydrogen atoms. 2 They include: n General/ Molecular formula Structural formula Name 1 CH4 H H C H H Methane 2 C2H6 H H H C C H H H Ethane 3 C3H8 H H H H C C C H H H H Propane 4 C4H10 H H H H H C C C C H H H H H Butane 5 C5H12 H H H H H H C C C C C H CH3 (CH2) 6CH3 H H H H H Pentane 6 C6H14 H H H H H H Hexane
3 H C C C C C C H CH3 (CH2) 6CH3 H H H H H H 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptane 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H H H Octane 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H H H Nonane 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H H H decane Note 1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g. Decane has a general/molecular formula C10H22 ;this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane.
The carbon atoms are linked by single bond to each other and to hydrogen atoms. 2 They include: n General/ Molecular formula Structural formula Name 1 CH4 H H C H H Methane 2 C2H6 H H H C C H H H Ethane 3 C3H8 H H H H C C C H H H H Propane 4 C4H10 H H H H H C C C C H H H H H Butane 5 C5H12 H H H H H H C C C C C H CH3 (CH2) 6CH3 H H H H H Pentane 6 C6H14 H H H H H H Hexane
3 H C C C C C C H CH3 (CH2) 6CH3 H H H H H H 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptane 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H H H Octane 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H H H Nonane 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H H H decane Note 1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g. Decane has a general/molecular formula C10H22 ;this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane. 2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom. The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms. 4 3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons.
2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom. The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms. 4 3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons. 4.Since Hydrogen is monovalent ,each atom of hydrogen in the alkane MUST always be bonded using one covalent bond/one shared pair of electrons. 5.One member of the alkane differ from the next/previous by a CH2 group. e.g Propane differ from ethane by one carbon and two Hydrogen atoms form ethane. Ethane differ from methane also by one carbon and two Hydrogen atoms 6.A group of compounds that differ by a CH2 group from the next /previous consecutively is called a homologous series. 7.A homologous series: (i) differ by a CH2 group from the next /previous consecutively (ii)have similar chemical properties (iii)have similar chemical formula that can be represented by a general formula e.g alkanes have the general formula CnH2n+2. (iv)the physical properties (e.g.melting/boiling points)show steady gradual change) 8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms. 9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g Alkane name molecular structure CnH2n+2 Alkyl name Molecula structure CnH2n+1 methane CH4 methyl CH3 ethane CH3CH3 ethyl CH3 CH2 propane CH3 CH2 CH3 propyl CH3 CH2 CH2 butane CH3 CH2 CH2 CH3 butyl CH3 CH2 CH2 CH2 (b)Isomers of alkanes Isomers are compounds with the same molecular general formula but different molecular structural formula.
(iv)the physical properties (e.g.melting/boiling points)show steady gradual change) 8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms. 9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g Alkane name molecular structure CnH2n+2 Alkyl name Molecula structure CnH2n+1 methane CH4 methyl CH3 ethane CH3CH3 ethyl CH3 CH2 propane CH3 CH2 CH3 propyl CH3 CH2 CH2 butane CH3 CH2 CH2 CH3 butyl CH3 CH2 CH2 CH2 (b)Isomers of alkanes Isomers are compounds with the same molecular general formula but different molecular structural formula. Isomerism is the existence of a compounds having the same general/molecular formula but different structural formula. The 1st three alkanes do not form isomers.Isomers are named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming. The IUPAC system of nomenclature uses the following basic rules/guidelines: 1.Identify the longest continuous carbon chain to get/determine the parent alkane. 5 2.Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens 4.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of branches attached to the parent alkane. Practice on IUPAC nomenclature of alkanes (a)Draw the structure of: (i)2-methylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2.
Practice on IUPAC nomenclature of alkanes (a)Draw the structure of: (i)2-methylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2” 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e Position of the branch at carbon “2” Number of branches at carbon “1” Type of the branch “methyl” hence Molecular formula CH3 CH3 CH CH2 CH3 // CH3 CH (CH3 ) CH2CH3 Structural formula H H H H H C C C C H H H H H C H
6 H (ii)2,2-dimethylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2” 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e Position of the branch at carbon “2” Number of branches at carbon “2” Type of the branch two“methyl” hence Molecular formular CH3 CH3 C CH2 CH3 // CH3 C (CH3 )2 CH2CH3 CH3 Structural formula H H C H H H H H C C C C H H H H H C H H
7 (iii) 2,2,3-trimethylbutane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2 and 3” 3.
Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2 and 3” 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e Position of the branch at carbon “2 and 3” Number of branches at carbon “3” Type of the branch three “methyl” hence Molecular formular CH3 CH3 C CH CH3 // CH3 C (CH3 )3 CH2CH3 CH3 CH3 Structural formula H H C H H H H C C C H H H H H C C H H H C H
8 H (iv) 1,1,1,2,2,2-hexabromoethane Molecular formula CBr3 CBr3 Structural formula Br Br Br C C Br Br Br (v) 1,1,1-tetrachloro-2,2-dimethylbutane CH3 CCl 3 C CH3 // C Cl 3 C (CH3 )2 CH3 CH3 Structural formula Cl Cl C Cl H H H C C C H H H H C H H (c)Occurrence and extraction Crude oil ,natural gas and biogas are the main sources of alkanes: (i)Natural gas is found on top of crude oil deposits and consists mainly of methane. 9 (ii)Biogas is formed from the decay of waste organic products like animal dung and cellulose. When the decay takes place in absence of oxygen , 60-75% by volume of the gaseous mixture of methane gas is produced. (iii)Crude oil is a mixture of many flammable hydrocarbons/substances. Using fractional distillation, each hydrocarbon fraction can be separated from the other. The hydrocarbon with lower /smaller number of carbon atoms in the chain have lower boiling point and thus collected first. As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point.
As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point. Uses of different crude oil fractions Carbon atoms in a molecule Common name of fraction Uses of fraction 1-4 Gas L.P.G gas for domestic use 5-12 Petrol Fuel for petrol engines 9-16 Kerosene/Paraffin Jet fuel and domestic lighting/cooking 15-18 Light diesel Heavy diesel engine fuel 18-25 Diesel oil Light diesel engine fuel 20-70 Lubricating oil Lubricating oil to reduce friction. Over 70 Bitumen/Asphalt Tarmacking roads (d)School laboratory preparation of alkanes In a school laboratory, alkanes may be prepared from the reaction of a sodium alkanoate with solid sodium hydroxide/soda lime. Chemical equation: Sodium alkanoate + soda lime -> alkane + Sodium carbonate CnH2n+1COONa(s) + NaOH(s) -> C n H2n+2 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CnH2n+1 in CnH2n+1COONa(s) to form C n H2n+2. 10 Examples 1. Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water Sodium ethanoate + soda lime -> methane + Sodium carbonate CH3COONa(s) + NaOH(s) -> C H4 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4. 2. Ethane is prepared from the heating of a mixture of sodium propanoate and soda lime and collecting over water Sodium propanoate + soda lime -> ethane + Sodium carbonate CH3 CH2COONa(s) + NaOH(s) -> CH3 CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 CH2 in CH3 CH2COONa (s) to form CH3 CH3 3.
Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water Sodium ethanoate + soda lime -> methane + Sodium carbonate CH3COONa(s) + NaOH(s) -> C H4 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4. 2. Ethane is prepared from the heating of a mixture of sodium propanoate and soda lime and collecting over water Sodium propanoate + soda lime -> ethane + Sodium carbonate CH3 CH2COONa(s) + NaOH(s) -> CH3 CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 CH2 in CH3 CH2COONa (s) to form CH3 CH3 3. Propane is prepared from the heating of a mixture of sodium butanoate and soda lime and collecting over water Sodium butanoate + soda lime -> propane + Sodium carbonate CH3 CH2CH2COONa(s) + NaOH(s) -> CH3 CH2CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 CH2 CH2 in CH3 CH2CH2COONa (s) to form CH3 CH2CH3 4. Butane is prepared from the heating of a mixture of sodium pentanoate and soda lime and collecting over water Sodium pentanoate + soda lime -> butane + Sodium carbonate CH3 CH2 CH2CH2COONa(s)+NaOH(s) -> CH3 CH2CH2CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3CH2 CH2 CH2 in CH3 CH2CH2 CH2COONa (s) to form CH3 CH2 CH2CH3 Laboratory set up for the preparation of alkanes
11 (d)Properties of alkanes I. Physical properties Alkanes are colourless gases, solids and liquids that are not poisonous. They are slightly soluble in water. The solubility decrease as the carbon chain and thus the molar mass increase The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase.
They are slightly soluble in water. The solubility decrease as the carbon chain and thus the molar mass increase The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkanes (methane,ethane,propane and butane)are therefore gases ,the nect six(pentane ,hexane, heptane,octane,nonane, and decane) are liquids while the rest from unidecane(11 carbon atoms) are solids . The density of straight chain alkanes increase with increasing carbon chain as the intermolecular forces increases. This reduces the volume occupied by a given mass of the compound. Summary of physical properties of alkanes Alkane General formula Melting point(K) Boiling point(K) Density gcm-3 State at room(298K) temperature and pressure atmosphere (101300Pa) Methane CH4 90 112 0.424 gas
12 Ethane CH3CH3 91 184 0.546 gas Propane CH3CH2CH3 105 231 0.501 gas Butane CH3(CH2)2CH3 138 275 0.579 gas Pentane CH3(CH2)3CH3 143 309 0.626 liquid Hexane CH3(CH2)4CH3 178 342 0.657 liquid Heptane CH3(CH2)5CH3 182 372 0.684 liquid Octane CH3(CH2)6CH3 216 399 0.703 liquid Nonane CH3(CH2)7CH3 219 424 0.708 liquid Octane CH3(CH2)8CH3 243 447 0.730 liquid II.Chemical properties (i)Burning. Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water. Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water.
Summary of physical properties of alkanes Alkane General formula Melting point(K) Boiling point(K) Density gcm-3 State at room(298K) temperature and pressure atmosphere (101300Pa) Methane CH4 90 112 0.424 gas
12 Ethane CH3CH3 91 184 0.546 gas Propane CH3CH2CH3 105 231 0.501 gas Butane CH3(CH2)2CH3 138 275 0.579 gas Pentane CH3(CH2)3CH3 143 309 0.626 liquid Hexane CH3(CH2)4CH3 178 342 0.657 liquid Heptane CH3(CH2)5CH3 182 372 0.684 liquid Octane CH3(CH2)6CH3 216 399 0.703 liquid Nonane CH3(CH2)7CH3 219 424 0.708 liquid Octane CH3(CH2)8CH3 243 447 0.730 liquid II.Chemical properties (i)Burning. Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water. Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water. Alkane + Air -> carbon(II) oxide + water (limited air) Examples 1.(a) Methane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Methane + Air -> carbon(IV) oxide + water (excess air/oxygen) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l/g) (b) Methane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.
Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water. Alkane + Air -> carbon(II) oxide + water (limited air) Examples 1.(a) Methane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Methane + Air -> carbon(IV) oxide + water (excess air/oxygen) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l/g) (b) Methane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. Methane + Air -> carbon(II) oxide + water (excess air/oxygen) 2CH4(g) + 3O2(g) -> 2CO(g) + 4H2O(l/g) 2.(a) Ethane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Ethane + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. 13 Ethane + Air -> carbon(II) oxide + water (excess air/oxygen) 2C2H6(g) + 5O2(g) -> 4CO(g) + 6H2O(l/g) 3.(a) Propane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Propane + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.