Đáp án:

\(\begin{array}{l} b,\ m_{\text{dd H$_2$SO$_4$}}=200\ g.\\ c,\ C\%_{ZnSO_4}=14,89\%\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\ PTHH:ZnO+H_2SO_4\to ZnSO_4+H_2O\\ b,\ n_{ZnO}=\dfrac{16,2}{81}=0,2\ mol.\\ Theo\ pt:\ n_{H_2SO_4}=n_{ZnSO_4}=n_{ZnO}=0,2\ mol.\\ \Rightarrow m_{\text{dd H$_2$SO$_4$}}=\dfrac{0,2\times 98}{9,8\%}=200\ g.\\ c,\ m_{\text{dd spư}}=m_{ZnO}+m_{\text{dd H$_2$SO$_4$}}=16,2+200=216,2\ g.\\ \Rightarrow C\%_{ZnSO_4}=\dfrac{0,2\times 161}{216,2}\times 100\%=14,89\%\end{array}\)

chúc bạn học tốt!

a,$ZnO+H_2SO_4→ZnSO_4+H_2O$

b,$n_{ZnO}=16,2/81=0,2(mol)$

$⇒n_{H_2SO_4}=0,2(mol)$

$⇒m_{H_2SO_4}=0,2.98=19,6(g)$

$⇒m_{ddH_2SO_4}=19,6.100/9,8=200(g)$

c,$n_{ZnSO_4}=0,2(mol)$
$⇒m_{ZnSO_4}=0,2.161=32,2(g)$

$m_{ddspu}=m_{ZnO}+m_{ddH_2SO_4}=16,2+200=216,2(g)$

$⇒$%$ZnSO_4=32,2.100/216,2≈14,9$%