$CuO+H_2SO_4\to CuSO_4+H_2O$
a/ $n_{CuO}=\dfrac{32}{80}=0,4(mol)$
Theo phương trình: $n_{CuO}=n_{H_2SO_4}$
$→n_{H_2SO_4}=0,4(mol)\\→m_{H_2SO_4}=0,4.98=39,2(g)\\m_{dd\,H_2SO_4}=\dfrac{39,2}{20\%}=196(g)$
Vậy $m=196g$
b/ Theo phương trình: $n_{CuO}=n_{CuSO_4}=n_{H_2O}$
$→n_{CuSO_4}=n_{H_2O}=0,4(mol)\\→\begin{cases}m_{CuSO_4}=0,4.160=64(g)\\m_{H_2O}=0,4.18=7,2(g)\end{cases}$
Vậy $m_{CuSO_4}=64g,\,m_{H_2O}=7,2g$
