\(n_S^{} = \frac{{6,4}}{{32}} = 0,2\;{\text{mol}}\)
\({\text{n}}_{Zn}^{} = \frac{{15}}{{65}} = \frac{3}{{13}}\;{\text{mol}}\)
\(Zn + S\xrightarrow{{t_{}^o}}ZnS\)
\(\frac{3}{{13}} > 0,2 \to Zn\;{\text{du}}{\text{, tinh theo S}}\)
\(0,2{\text{                  0}}{\text{,2}}\)
\(m_{ZnS}^{} = 0,2.97 = 19,4g\)
\(m_{Zn_{Du}^{}}^{} = 65.(\frac{3}{{13}} - 0,2) = 2g\)

Đáp án:

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Giải thích các bước giải:

$\begin{array}{1}n_{Zn}=\dfrac{15}{65}=\dfrac{3}{13}(mol)\\n_{S}=\dfrac{6,4}{32}=0,2(mol)\\PTHH:Zn+S \xrightarrow{t^o} ZnS\\\to \text{Zn dư sau phản ứng}\to n_{Zn}(dư)=\dfrac{3}{13}-0,2=\dfrac{2}{65}(mol)\\\to m_{Zn}(du)=\dfrac{2}{65} \times 65=2(g)\\n_{ZnS}=n_{S}=0,2(mol)\\\to m_{ZnS}=0,2 \times 97=19,4(g)\end{array}$

`text{# Như}`