5)
Phản ứng xảy ra:
\(FeO + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}O\)
Ta có:
\({n_{FeO}} = \frac{{16,2}}{{56 + 16}} = 0,225{\text{ mol}}\)
\( \to {n_{HCl}} = 2{n_{FeO}} = 0,225.2 = 0,45{\text{ mol}}\)
\( \to {m_{HCl}} = 0,45.36,5 = 16,425{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{16,425}}{{200}}.100\% = 8,2125\% \)
BTKL:
\({m_{dd}} = {m_{FeO}} + {m_{dd\;{\text{HCl}}}} = 16,2 + 200 = 216,2{\text{ gam}}\)
\({n_{FeC{l_2}}} = {n_{FeO}} = 0,225{\text{ mol}}\)
\( \to {m_{FeC{l_2}}} = 0,225.(56 + 35,5.2) = 28,575{\text{ gam}}\)
\( \to C{\% _{FeC{l_2}}} = \frac{{28,575}}{{216,2}} = 13,2\% \)
6)
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol}}\)
\( \to {n_{{H_2}}} = {n_{Zn}} = 0,2{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
\({n_{ZnC{l_2}}} = {n_{Zn}} = 0,2{\text{ mol}}\)
\( \to {m_{ZnC{l_2}}} = 0,2.(65 + 35,5.2) = 27,2{\text{ gam}}\)
\({m_{dd}} = {m_{Zn}} + {m_{dd{\text{ }}{\text{HCl}}}} - {m_{{H_2}}} = 13 + 200 - 0,2.2 = 212,6{\text{ gam}}\)
\( \to C{\% _{ZnC{l_2}}} = \frac{{27,2}}{{212,6}} = 12,8\% \)
\({n_{HCl}} = 2{n_{Zn}} = 0,2.2 = 0,4{\text{ mol}}\)
\( \to {m_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{14,6}}{{200}} = 7,3\% \)
8)
Phản ứng xảy ra:
\(CuO + 2HCl\xrightarrow{{}}CuC{l_2} + {H_2}O\)
Ta có:
\({n_{CuO}} = \frac{{1,6}}{{64 + 16}} = 0,02{\text{ mol}}\)
\({m_{HCl}} = 100.3,65\% = 3,65{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{{3,65}}{{36,5}} = 0,1{\text{ mol > 2}}{{\text{n}}_{CuO}}\)
Vậy \(HCl\) dư
\({n_{HCl{\text{ dư}}}} = 0,1 - 0,02.2 = 0,06{\text{ mol}}\)
\((\to {m_{HCl \text{ dư}}} = 0,06.36,5 = 2,19{\text{ gam}}\)
\({n_{CuC{l_2}}} = {n_{CuO}} = 0,02{\text{ mol}}\)
\( \to {m_{CuC{l_2}}} = 0,02.(64 + 35,5.2) = 2,7{\text{ gam}}\)
BTKL:
\({m_{dd}} = {m_{CuO}} + {m_{dd\;{\text{HCl}}}} = 1,6 + 100 = 101,6{\text{ gam}}\)
\( \to C{\% _{CuC{l_2}}} = \frac{{2,7}}{{101,6}} = 2,66\% \)
\(C{\% _{HCl}} = \frac{{2,19}}{{101,6}} = 2,15\% \)
