Bài 1.
a,
$CuO+H_2→Cu+H_2O$
$n_{H_2}=\frac{4,48}{22,4}=0,2mol$
$→n_{CuO}=0,2mol$
$→m_{CuO}=0,2.80=16g$
b,
$n_{Cu}=H_2=0,2mol$
$→m_{Cu}=0,2.64=12,8g$
Bài 2.
$n_{O_2}=\frac{0,56.\frac{1}{5}}{22,4}=0,005mol$
a,
$S+O_2→SO_2$
$→n_S=0,005mol$
$→m_S=0,005.32=0,16g$
b,
$n_{SO_2}=n_{O_2}=0,005mol$
$→V_{SO_2}=0,005.22,4=0,112l$