\(n_{Zn}^{} = \frac{m}{M} = \frac{{3,5}}{{65}} = \frac{7}{{130}}{\text{ mol}}\)

\(a)Zn + 2HCl \to ZnCl_2^{} + H_2^{}\)

\(b)\frac{7}{{130}}{\text{    }}\frac{7}{{65}}{\text{    }}\frac{7}{{130}}{\text{     }}\frac{7}{{130}}\)

\(V_{H_2^{}}^{} = \frac{7}{{130}}.22,4 = 1,206l\)

\(m_{HCl}^{} = \frac{7}{{65}}.36,5 = 3,93g\)

\(c)m_{{\text{dd}}HCl}^{} = 3,93.\frac{{100}}{{30}} = 13,1g\)

\(m_{ZnCl_2^{}}^{} = \frac{7}{{130}}.136 = 7,32g\)

\(BTKL:m_{Zn}^{} + m_{HCl}^{} = m_{{\text{ddSauPu}}}^{} + m_{H_2^{}}^{}\)

\(m_{{\text{dd}}_{sauPU}^{}}^{} = 3,5 + 13,1 - \frac{7}{{130}}.2 = 16,5g\)

\(C\% _{ZnCl_2^{}}^{} = \frac{{7,32}}{{16,5}}.100 = 44,32\% \)

Đáp án + Giải thích các bước giải:

$a\big)Zn+2HCl\to ZnCl_2+H_2$

$b\big)$

$n_{Zn}=\dfrac{3,5}{65}=\dfrac{7}{130}(mol)$

Theo PT: $n_{H_2}=n_{Zn}=\dfrac{7}{130}(mol)$

$\to V_{H_2}=\dfrac{7}{130}.22,4≈1,2(l)$

$c\big)$

Theo PT: $n_{HCl}=2n_{Zn}=\dfrac{7}{65}(mol)$

$\to m_{dd\,HCl}=\dfrac{\frac{7}{65}.36,5}{30\%}\approx 13,1(g)$

$d\big)$

Theo PT: $n_{ZnCl_2}=n_{Zn}=\dfrac{7}{130}(mol)$

$\to m_{ZnCl_2}=\dfrac{7}{130}.136≈7,32(g)$

$e\big)$

$m_{dd\,spu}=3,5+13,1-\dfrac{7}{130}.2\approx 16,49(g)$

$\to C\%_{ZnCl_2}=\dfrac{7,32}{16,49}.100\%≈44,39\%$