6)
Phản ứng xảy ra:
\(2{K_3}P{O_4} + 3Mg{(N{O_3})_2}\xrightarrow{{}}M{g_3}{(P{O_4})_2} \downarrow + 6KN{O_3}\)
Ta có:
\({n_{{K_3}P{O_4}}} = 0,15.4 = 0,6{\text{ mol;}}{{\text{n}}_{Mg{{(N{O_3})}_2}}} = 0,25.3,8 = 0,95{\text{ mol > }}\frac{3}{2}{n_{{K_3}P{O_4}}} \to Mg{(N{O_3})_2}\) dư.
\( \to {n_{Mg{{(N{O_3})}_2}{\text{ dư}}}} = 0,95 - \frac{3}{2}{n_{{K_3}P{O_4}}} = 0,05{\text{ mol;}}{{\text{n}}_{M{g_3}{{(P{O_4})}_2}}} = \frac{1}{2}{n_{{K_3}P{O_4}}} = 0,3{\text{ mol;}}{{\text{n}}_{KN{O_3}}} = 3{n_{{K_3}P{O_4}}} = 1,8{\text{ mol}}\)
\( \to {m_{Mg{{(N{O_3})}_2}{\text{ dư}}}} = 0,05.(24 + 62.2) = 7,4{\text{ gam;}}{{\text{m}}_{M{g_3}{{(P{O_4})}_2}}} = 0,3.(24.3 + 95.2) = 78,6{\text{ gam}}\)
\({V_{dd{\text{ sau phản ứng}}}} = 150 + 250 = 400ml = 0,4{\text{ lít}}\)
\( \to {C_{M{\text{ Mg(N}}{{\text{O}}_3}{)_2}}} = \frac{{0,05}}{{0,4}} = 0,125M;{C_{M{\text{ }}{\text{KN}}{{\text{O}}_3}}} = \frac{{1,8}}{{0,4}} = 4,5M\)
7)
Phản ứng xảy ra:
\(Pb{(N{O_3})_2} + 2HCl\xrightarrow{{}}PbC{l_2} + 2HN{O_3}\)
Ta có:
\({n_{HCl}} = 0,2.0,2 = 0,4{\text{ mol;}}{{\text{n}}_{Pb{{(N{O_3})}_2}}} = 0,1.1,5 = 0,15{\text{ mol < }}\frac{1}{2}{n_{HCl}}\)
Vậy HCl dư
\({n_{PbC{l_2}}} = {n_{Pb{{(N{O_3})}_2}}} = 0,15{\text{ mol}} \to {{\text{m}}_{PbC{l_2}}} = 0,15.(207 + 35,5.2) = 41,7{\text{ gam}}\)
\({n_{HCl{\text{ dư}}}} = 0,4 - 2{n_{Pb{{(N{O_3})}_2}}} = 0,1{\text{ mol;}}{{\text{n}}_{HN{O_3}}} = 2{n_{Pb{{(N{O_3})}_2}}} = 0,3{\text{ mol}}\)
\({V_{dd}} = 200 + 100 = 300ml = 0,3{\text{ lít}}\)
\({C_{M{\text{ HCl}}}} = \frac{{0,1}}{{0,3}} = 0,333M;{C_{M{\text{ HN}}{{\text{O}}_3}}} = \frac{{0,3}}{{0,3}} = 1M\)
