\(\begin{array}{l}
a)CuO+H_2SO_4\to CuSO_4+H_2O\\
b)\\
n_{CuO}=\frac{1,6}{80}=0,02(mol)\\
n_{H_2SO_4}=\frac{100.20\%}{98}\approx 0,2(mol)\\
Vì\,n_{CuO}<n_{H_2SO_4}\to H_2SO_4\,\rm dư\\
Theo\,PT:\,n_{CuSO_4}=n_{H_2SO_4(pu)}=0,02(mol)\\
\to m_{CuSO_4(du)}=100.20\%-0,02.98=18,04(g)\\
m_{dd\,spu}=1,6+100=101,6(g)\\
\to\begin{cases} C\%_{CuSO_4}=\frac{0,02.160}{101,6}.100\%≈3,15\%\\ C\%_{H_2SO_4(du)}=\frac{18,04}{101,6}.100\%≈17,76\%\end{cases}
\end{array}\)

Đáp án:

$a,\ CuO+H_2SO_4\to CuSO_4+H_2O\\ b,\\ \Rightarrow C\%_{dd\ CuSO_4}=3,15\%\\ \Rightarrow C\%_{dd\ H_2SO_4\ dư}=17,76\%$

Giải thích các bước giải:

$a,\\ CuO+H_2SO_4\to CuSO_4+H_2O\ (*)\\ b,\\ n_{CuO}=\dfrac{1,6}{80}=0,02(mol)\\ n_{H_2SO_4}=\dfrac{100.20\%}{98}=\dfrac{10}{49}(mol)\\ Xét\:\ \dfrac{n_{CuO}}{1}<\dfrac{n_{H_2SO_4}}{1}\\ \to H_2SO_4\ dư\\ \xrightarrow{\text{Từ PTHH}}\ n_{CuSO_4}=n_{H_2SO_4\ (*)}=n_{CuO}=0,02(mol)\\ n_{H_2SO_4\ dư}=\dfrac{10}{49}-0,02=\dfrac{451}{2450}(mol)\\ m_{CuSO_4}=0,02.160=3,2(g)\\ m_{H_2SO_4\ dư}=\dfrac{451}{2450}.98=18,04(g)\\ m_{dd\ sau}=1,6+100=101,6(g)\\$

$\Rightarrow C\%_{dd\ CuSO_4}=\dfrac{3.2}{101,6}.100\%=3,15\%\\ \Rightarrow C\%_{dd\ H_2SO_4\ dư}=\dfrac{18,04}{101,6}.100\%=17,76\%$