дано
m(CuO) = 80 g
V(H2) -?
CuO+H2--->Cu+H2O
M(CuO ) = 80 g/mol
n(CuO) = m/M = 80 / 80 mol
n(CuO) = n(H2) = 1 mol
V(H2) = n(H2) * Vm = 1*22.4 = 22.4 L
ответ 22.4 л
дано
m(CuO) = 80 g
V(H2) -?
CuO+H2--->Cu+H2O
M(CuO ) = 80 g/mol
n(CuO) = m/M = 80 / 80 mol
n(CuO) = n(H2) = 1 mol
V(H2) = n(H2) * Vm = 1*22.4 = 22.4 L
ответ 22.4 л