1)
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
\(CuO + 2HCl\xrightarrow{{}}CuC{l_2} + {H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol = }}{{\text{n}}_{Fe}}\)
Ta có:
\({m_{HCl}} = 219.10\% = 21,9{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{{21,9}}{{36,5}} = 0,6{\text{ mol = 2}}{{\text{n}}_{Fe}} + 2{n_{CuO}}\)
\( \to {n_{CuO}} = \frac{{0,6 - 0,2.2}}{2} = 0,1{\text{ mol}}\)
\( \to {m_{Fe}} = 0,2.56 = 11,2{\text{ gam;}}{{\text{m}}_{CuO}} = 0,1.(64 + 16) = 8{\text{ gam}}\)
\( \to \% {m_{Fe}} = \frac{{11,2}}{{11,2 + 8}}.100\% = 58,33\% \)
2)
Phản ứng xảy ra:
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{NaOH}} = \frac{8}{{40}} = 0,2{\text{ mol}}\)
\( \to {n_{{H_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = 0,1{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,1.98 = 9,8{\text{ gam}}\)
3)
Phản ứng xảy ra:
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có:
\({m_{{H_2}S{O_4}}} = 375.9,8\% = 36,75{\text{ gam}}\)
\( \to {n_{{H_2}S{O_4}}} = \frac{{36,75}}{{98}} = 0,375{\text{ mol = }}{{\text{n}}_{N{a_2}S{O_4}}}\)
\( \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,375.2 = 0,75{\text{ mol}}\)
\( \to {m_{NaOH}} = 0,75.40 = 30{\text{ gam}}\)
\( \to {m_{dd{\text{ NaOH}}}} = \frac{{30}}{{20\% }} = 150{\text{ gam}}\)
\( \to {m_{dd{\text{ B}}}} = 375 + 150 = 525{\text{ gam}}\)
\({m_{N{a_2}S{O_4}}} = 0,375.(23.2 + 96) = 53,25{\text{ gam}}\)
\( \to C{\% _{N{a_2}S{O_4}}} = \frac{{53,25}}{{525}}.100\% = 10,14\% \)
