3)
дано
n(Cu) = 0.4 mol
----------------
V(O2)-?
2Cu+O2-->2CuO
2n(Cu) = n(O2)
n(O2) = 0.4 / 2 = 0.2 mol
V(O2) = n(O2) * Vm = 0.2*22.4 = 4.48 L
ответ 4.48 л
4)
ДАНО
m(Cl2) = 70 g
-------------------
m(Fe)-?
2Fe+3CL2-->2FeCL3
M(CL2) = 71 g/mol
n(CL2) = m/M = 70 / 71 = 1 mol
2n(Fe) = 3n(CL2)
n(Fe) = 2*1 / 3= 0.67 mol
M(Fe) = 56 g/mol
m(Fe) = n*M = 0.67 * 56 = 37.52 g
ответ 37.52 г