дано
m(Ag2O) = 116 g
V(O2)-?
4Ag+O2-->2Ag2O
M(Ag2O) = 232 g/mol
n(Ag2O) = m/M = 116 / 232 = 0.5 mol
n(O2) = 2n(Ag2O)
n(O2) = 0.5 / 2 = 0.25 mol
V(O2) = n*Vm = 0.25 * 22.4 = 5.6 L
ответ 5.6 л
дано
m(Ag2O) = 116 g
V(O2)-?
4Ag+O2-->2Ag2O
M(Ag2O) = 232 g/mol
n(Ag2O) = m/M = 116 / 232 = 0.5 mol
n(O2) = 2n(Ag2O)
n(O2) = 0.5 / 2 = 0.25 mol
V(O2) = n*Vm = 0.25 * 22.4 = 5.6 L
ответ 5.6 л