Câu 10:

a,

$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$

$A+2HCl\to ACl_2+H_2$

$\Rightarrow n_A=0,1(mol)$

$M_A=\dfrac{6,5}{0,1}=65(Zn)$

$\to A$ là kẽm 

b,

$n_{HCl}=2n_{H_2}=0,2(mol)$

$\Rightarrow m_{dd HCl}=0,2.36,5:7,3\%=100g$

$\to V_{dd}=\dfrac{100}{1,095}=91,3ml$

Đáp án:

\(\begin{array}{l} a,\ \text{A là Kẽm (Zn).}\\ b,\ V_{\text{dd HCl}}=91,32\ ml.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\\ PTHH:A+2HCl\to ACl_2+H_2↑\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ Theo\ pt:\ n_{A}=n_{H_2}=0,1\ mol.\\ \Rightarrow M_{A}=\dfrac{6,5}{0,1}=65\ (g/mol)\\ \Rightarrow \text{A là Kẽm (Zn).}\\ b,\\ PTHH:Zn+2HCl\to ZnCl_2+H_2↑\\ Theo\ pt:\ n_{HCl}=2n_{H_2}=0,2\ mol.\\ \Rightarrow m_{\text{dd HCl}}=\dfrac{0,2\times 36,5}{7,3\%}=100\ g.\\ \Rightarrow V_{\text{dd HCl}}=\dfrac{100}{1,095}=91,32\ ml.\end{array}\)

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