1)
Phản ứng xảy ra:
\(2Ba + {O_2}\xrightarrow{{{t^o}}}2BaO\)
Ta có:
\({n_{Ba}} = \frac{{13,7}}{{137}} = 0,1{\text{ mol;}}{{\text{n}}_{{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol > }}\frac{1}{2}{n_{Ba}}\)
Vậy \(O_2\) dư
\( \to {n_{{O_2}{\text{ dư}}}} = 0,2 - \frac{1}{2}.0,1 = 0,15{\text{ mol}}\)
\( \to {m_{{O_2}{\text{ dư}}}} = 0,15.32 = 4,8{\text{ gam}}\)
2)
Phản ứng xảy ra:
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
Ta có:
\({n_{Al}} = \frac{{6,75}}{{27}} = 0,25{\text{ mol}}\)
\({m_{{H_2}S{O_4}}} = 98.30\% = 29,4{\text{ gam}} \to {{\text{n}}_{{H_2}S{O_4}}} = \frac{{29,4}}{{98}} = 0,3{\text{ mol}}\)
\( \to {n_{Al}} > \frac{2}{3}{n_{{H_2}S{O_4}}}\) nên \(Al\) dư.
\( \to {n_{Al{\text{ dư}}}} = 0,25 - \frac{2}{3}.0,3 = 0,05{\text{ mol}}\)
\( \to {m_{Al{\text{ dư}}}} = 0,05.27 = 1,35{\text{ gam}}\)
3)
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{Fe}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol;}}{{\text{n}}_{HCl}} = 0,1.3 = 0,3{\text{ mol}}\)
\( \to {n_{HCl}} > 2{n_{Fe}}\) nên \(HCl\) dư
\( \to {m_{HCl{\text{ dư}}}} = 0,3 - 0,1.2 = 0,1{\text{ mol}}\)
\( \to {m_{HCl{\text{ dư}}}} = 0,1.36,5 = 3,65{\text{ gam}}\)
4)
Phản ứng xảy ra:
\(2KOH + {H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O\)
Ta có:
\({m_{{H_2}S{O_4}}} = 200.9,8\% = 19,6{\text{ gam}} \to {n_{{H_2}S{O_4}}} = \frac{{19,6}}{{98}} = 0,2{\text{ mol}}\)
\({m_{KOH}} = 200.5,6\% = 11,2{\text{ gam}}\)
\( \to {n_{KOH}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol}}\)
\( \to {n_{{H_2}S{O_4}}} > \frac{1}{2}{n_{KOH}}\)
Vậy \(H_2SO_4\) dư
\( \to {n_{{K_2}S{O_4}}} = \frac{1}{2}{n_{KOH}} = 0,1{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}{\text{ dư}}}} = 0,2 - \frac{1}{2}.0,2 = 0,1{\text{ mol}}\)
\( \to {m_{{K_2}S{O_4}}} = 0,1.(39.2 + 32 + 16.4) = 16{\text{ gam}}\)
\({m_{{H_2}S{O_4}{\text{ dư}}}} = 0,1.98 = 9,8{\text{ gam}}\)
\({m_{dd}} = {m_{dd\;{{\text{H}}_2}S{O_4}}} + {m_{dd\;{\text{KOH}}}} = 200 + 200 = 400{\text{ gam}}\)
\( \to C{\% _{{H_2}S{O_4}}} = \frac{{9,8}}{{400}}.100\% = 2,45\% ;C{\% _{{K_2}S{O_4}}} = \frac{{16}}{{400}}.100\% = 4\% \)
5)
Phản ứng xảy ra:
\(HN{O_3} + NaOH\xrightarrow{{}}NaN{O_3}\)
Ta có:
\({n_{HN{O_3}}} = 0,15.2 = 0,3{\text{ mol = }}{{\text{n}}_{NaOH}} = {n_{NaN{O_3}}}\)
\( \to {V_{dd\;{\text{NaOH}}}} = \frac{{0,3}}{2} = 0,15{\text{ lít}}\)
\( \to {V_{dd}} = 0,15 + 0,15 = 0,3{\text{ lít}}\)
\( \to {C_{M{\text{ NaN}}{{\text{O}}_3}}} = \frac{{0,3}}{{0,3}} = 1M\)
