дано
m техн(C) = 80 g
V(CO2) = 112 L
---------------------
W(прим)-?
C+O2-->CO2
n(CO2) = V(CO2) / Vm = 112 / 22.4 = 5 mol
n(C) = n(CO2) = 5 mol
M(C) = 12 g/mol
m(C)= n*M = 5*12 = 60 g
W(прим) = (80 - 60) / 80 * 100% = 25%
ответ 25%
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Verified answer
дано
m техн(C) = 80 g
V(CO2) = 112 L
---------------------
W(прим)-?
C+O2-->CO2
n(CO2) = V(CO2) / Vm = 112 / 22.4 = 5 mol
n(C) = n(CO2) = 5 mol
M(C) = 12 g/mol
m(C)= n*M = 5*12 = 60 g
W(прим) = (80 - 60) / 80 * 100% = 25%
ответ 25%