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:

\({m_{{H_2}S{O_4}}} = 375.9,8\%  = 36,75{\text{ gam}}\)

\( \to {n_{{H_2}S{O_4}}} = \frac{{36,75}}{{98}} = 0,375{\text{ mol = }}{{\text{n}}_{N{a_2}S{O_4}}}\)

\( \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,375.2 = 0,75{\text{ mol}}\)

\( \to {m_{NaOH}} = 0,75.40 = 30{\text{ gam}}\)

\( \to {m_{dd{\text{ NaOH}}}} = \frac{{30}}{{20\% }} = 150{\text{ gam}}\)

\( \to {m_{dd{\text{ B}}}} = 375 + 150 = 525{\text{ gam}}\)

\({m_{N{a_2}S{O_4}}} = 0,375.(23.2 + 96) = 53,25{\text{ gam}}\)

\( \to C{\% _{N{a_2}S{O_4}}} = \frac{{53,25}}{{525}}.100\%  = 10,14\% \)

1) Fe + 2HCl => FeCl2 + H2

CuO + 2HCl => CuCl2 + H2O

nH2 = 0,2 mol = nFe.

nHCl = 219 x 10/(100 x 36,5) = 0.6 mol

nFe + nCuO = 1/2nHCl =>n CuO = 0,3 - 0,2 = 0,1 mol

% Fe = 56 x 0,2 / (56 x 0,2 + 80 x 0,1 ) = 58.3 %

2)

H2SO4 + 2NaOH => Na2SO4 + H2O

nH2SO4 =1/2n NaOH = 1/2 x 8/40 = 0,1 mol

mH2SO4 = 0,1 x 98 = 9,8 g

3)

H2SO4 + 2NaOH => Na2SO4 + H2O

nNaOH = 2nH2SO4 = 2 x 375 x 9,8/(100 x 98) = 0.75 mol

mNaOH = 0,75 x 40 = 30g

mdd NaOH = 30 x 100 / 20 = 150g

dd sau phản ứng có mdd: 150 + 375 = 525g

nNa2SO4 = 1/2nNaOH = 0,375 mol

mNa2SO4 = 0,375 x 142 = 53,25g

C % Na2SO4 = 53,25 x 100 / 525 =10.14%