дано
m(CxHy) = 4.3 g
V(CO2) = 6.72 L
m(H2O) = 6.3 g
D(H2) = 43
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CxHy-?
Решаю согласно данному алгоритму
1)
M(CxHy) = D(H2) * M(H2) = 43*2 = 86 g/mol
2)
m(C) = V(CO2)*M(C) / Vm
M(C) = 12 g/mol , Vm = 22.4 L/mol
m(C) = 6.72 * 12 / 22.4 = 3.6 g
m(H) = m(H2O) * m(H)*n(H) / M(H2O)
M(H2O) = 18 g/mol
m(H) = 1 g , n(H) = 2 mol
m(H) = 6.3 * 1*2 / 18 = 0.7 g
3)
W(C) = m(C) / m(CxHy) = 3.6 / 4.3 = 0.84
W(H) = 1-W(C) = 1-0.84 = 0.16
4)
n(C) = w(C) * M(CxHy) / M(C) = 0.84 * 86 / 12 = 6
n(H) = W(H) * M(CxHy) / M(H) = 0.16* 86 / 1 = 14
n(C) : n(H) = 6 : 14
C6H14
ответ ГЕКСАН