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for the complete combustion of 5.6dm3 of a gaseuos hydrocarbon CxHy. 28.0dm3 of oxygen gas were used, 16.8dm3 of CO2 gas and 18.0gof liquid water were produced all gases measurement were made at stp. Determine the chemical formula of the hydrocarbon

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Answer is: the chemical formula of the hydrocarbon is C₃H₈.
Chemical reaction: CxHy + O₂ → xCO₂ + y/2H₂O.
V(CxHy) = 5,6 dm³.
V(O₂) = 28 dm³.
V(CO₂) = 16,8 dm³.
m(H₂O) = 18 g.
Vm = 22,4 dm³/mol; molar volume.
n(CxHy) = V(CxHy) ÷ Vm.
n(CxHy) = 5,6 dm³ ÷ 22,4 dm³/mol = 0,25 mol.
n(O₂) = 28 dm³ ÷ 22,4 dm³/mol = 1,25 mol.
n(CO₂) = 16,8 dm³ ÷ 22,4 dm = 0,75 mol.
n(H₂O) = 18 g ÷ 18 g/mol = 1 mol.
n(CxHy) : n(O₂) : n(CO₂) : n(H₂O) = 0,25 mol : 1,25 mol : 0,75 mol : 1 mol. 
n(CxHy) : n(O₂) : n(CO₂) : n(H₂O) = 1 : 5 : 3 : 4.
Reaction is than: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
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