Đáp án:
\(m=43,05 \text{ gam}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
\(2{\text{A}}l + 6HCl\xrightarrow{{}}2{\text{A}}lC{l_3} + 3{H_2}\)
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
\(MgC{l_2} + 2{\text{A}}gN{O_3}\xrightarrow{{}}Mg{(N{O_3})_2} + 2{\text{A}}gCl\)
\(AlC{l_3} + 3{\text{A}}gN{O_3}\xrightarrow{{}}3{\text{A}}gCl + Al{(N{O_3})_3}\)
\(ZnC{l_2} + 2{\text{A}}gN{O_3}\xrightarrow{{}}Zn{(N{O_3})_2} + 2{\text{A}}gCl\)
Ta có:
\({m_{dd{\text{ tăng}}}} = {m_{kl}} - {m_{{H_2}}} \to 3,58 = 3,88 - {m_{{H_2}}}\)
\( \to {n_{{H_2}}} = \frac{{0,3}}{2} = 0,15{\text{ mol}} \to {{\text{n}}_{HCl}} = 2{n_{{H_2}}} = 0,3{\text{ mol}}\)
\( \to {n_{AgCl}} = {n_{HCl}} = 0,3{\text{ mol}}\)
\( \to m = {m_{AgCl}} = 0,3.143,5 = 43,05{\text{ gam}}\)