更新1:
the combustion reaction of a fuel with 9.1070g of oxygen yeilds 11.5590g of carbon dioxide and 2.3645g of water. determine the empirical formula.
how to calculate?help me!!!please!!quick!!!?
2015-09-23 9:41 pm
the combustion reaction of a fuel with 9.1070g of oxygen yeild 11.5590g of carbon dioxide and 2.365g of water. determine the empirical formula.
回答 (2)
2015-09-26 9:52 am
Molar mass of H = 1.008 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H2O = 1.008×2 + 16.00 = 18.016 g/mol
Molar mass of CO2 = 12.01 + 16.00×2 = 44.01 g/mol
Mass of C in the fuel
= Mass of C in CO2 formed
= 11.5590 × (12.01/44.01) g
= 3.154 g
Mass of H in the fuel
= Mass of H in H2O formed
= 2.365 × (1.008×2/18.016)
= 0.2646g
By the law of mass of conservation :
Mass of O in the fuel
= (Total mass of the products) - (Mass of O2) - (Mass of C and H in the fuel)
= (11.5590 + 2.365) - 9.1070 - (3.154 + 0.2646)
= 1.399 g
In the fuel, mole ratio C : H : O
= (3.154/12.01) : (0.2646/1.008) : (1.399/16.00)
= 0.2626 : 0.2625 : 0.08744
= 3 : 3 : 1
Empirical formula of the fuel = C3H3O ...... (Ans)
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H2O = 1.008×2 + 16.00 = 18.016 g/mol
Molar mass of CO2 = 12.01 + 16.00×2 = 44.01 g/mol
Mass of C in the fuel
= Mass of C in CO2 formed
= 11.5590 × (12.01/44.01) g
= 3.154 g
Mass of H in the fuel
= Mass of H in H2O formed
= 2.365 × (1.008×2/18.016)
= 0.2646g
By the law of mass of conservation :
Mass of O in the fuel
= (Total mass of the products) - (Mass of O2) - (Mass of C and H in the fuel)
= (11.5590 + 2.365) - 9.1070 - (3.154 + 0.2646)
= 1.399 g
In the fuel, mole ratio C : H : O
= (3.154/12.01) : (0.2646/1.008) : (1.399/16.00)
= 0.2626 : 0.2625 : 0.08744
= 3 : 3 : 1
Empirical formula of the fuel = C3H3O ...... (Ans)
2015-09-24 8:36 am
CH
收錄日期: 2021-04-18 00:16:53
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