дано
m(CH3NH2) = 93 g
V(CO2) =?
4CH3NH2+9O2-->4CO2+10H2O+2N2
M(CH3NH2) = 31 g/mol
n(CH3NH2) = m/M = 93 / 31 = 3 mol
4n(CH3NH2) = 4(CO2) = 3 mol
V(CO2) = n*Vn = 3*22.4 = 67.2 L
ответ 67.2 л
дано
m(CH3NH2) = 93 g
V(CO2) =?
4CH3NH2+9O2-->4CO2+10H2O+2N2
M(CH3NH2) = 31 g/mol
n(CH3NH2) = m/M = 93 / 31 = 3 mol
4n(CH3NH2) = 4(CO2) = 3 mol
V(CO2) = n*Vn = 3*22.4 = 67.2 L
ответ 67.2 л