Ответ:
дано
m гряз(C10H22 ) = 400 g
W(прим) = 10 %
-------------------
V(CO2) - ?
m чист(C10H22) = m(грязC10H22 ) - (m(гряз C10H22 )*W(прим) / 100% ) = 400 - (400*10% / 100%) = 360 g
2C10H22+31O2-->20CO2+22H2O
M(C10H22) = 142 g/mol
n(C10H22) = m/M = 360 / 142 = 2.54 mol
2n(C10H22) = 20 n(CO2)
n(CO2) = 2.54*20 / 2 = 25,4 mol
V(CO2) = n*Vm = 25.4 * 22.4 = 568.96 L
ответ 568.96 л
Объяснение:
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Ответ:
дано
m гряз(C10H22 ) = 400 g
W(прим) = 10 %
-------------------
V(CO2) - ?
m чист(C10H22) = m(грязC10H22 ) - (m(гряз C10H22 )*W(прим) / 100% ) = 400 - (400*10% / 100%) = 360 g
2C10H22+31O2-->20CO2+22H2O
M(C10H22) = 142 g/mol
n(C10H22) = m/M = 360 / 142 = 2.54 mol
2n(C10H22) = 20 n(CO2)
n(CO2) = 2.54*20 / 2 = 25,4 mol
V(CO2) = n*Vm = 25.4 * 22.4 = 568.96 L
ответ 568.96 л
Объяснение: