1)
Gọi số mol \(KMnO_4;KClO_3\) lần lượt là \(x;y\)
\( \to 158x + 122,5y = 56,1{\text{ gam}}\)
Phản ứng xảy ra:
\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)
Ta có:
\({n_{{O_2}}} = \frac{1}{2}{n_{KMn{O_4}}} + \frac{3}{2}{n_{KCl{O_3}}} = 0,5x + 1,5y = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol}}\)
Giải được:
\(x=y=0,2\)
\( \to {m_{KMn{O_4}}} = 0,2.(39 + 55 + 16.4) = 31,6{\text{ gam}}\)
\( \to \% {m_{KMn{O_4}}} = \frac{{31,6}}{{56,1}}.100\% = 56,33\% \to \% {m_{KCl{O_3}}} = 43,67\% \)
3)
Gọi số mol \(Mg;Al\) lần lượt là \(x;y\)
\( \to 24x + 27y = 7,8{\text{ gam}}\)
Phản ứng xảy ra:
\(2Mg + {O_2}\xrightarrow{{{t^o}}}2MgO\)
\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)
Ta có:
\({n_{{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol = }}\frac{1}{2}{n_{Mg}} + \frac{3}{4}{n_{Al}} = 0,5x + 0,75y = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)
Giải được:
\(x=0,1;y=0,2\)
\( \to {m_{Mg}} = 0,1.24 = 2,4{\text{ gam}}\)
\( \to \% {m_{Mg}} = \frac{{2,4}}{{7,8}}.100\% = 30,77\% \to \% {m_{Al}} = 69,23\% \)
4)
Gọi số mol \(Fe;Al\) lần lượt là \(x;y\)
\( \to 56x + 27y = 11{\text{ gam}}\)
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
Ta có:
\({n_{{H_2}S{O_4}}} = 0,4.1 = 0,4{\text{ mol = }}{{\text{n}}_{Fe}} + \frac{3}{2}{n_{Al}}\)
Giải được: \(x=0,1;y=0,2\)
\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ }}gam \to {m_{Al}} = 11 - 5,6 = 5,4{\text{ gam}}\)
\({n_{FeS{O_4}}} = {n_{Fe}} = 0,1{\text{ mol;}}{{\text{n}}_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,1{\text{ mol}}\)
\({V_{dd}} = {V_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = 400{\text{ ml = 0}}{\text{,4 lít}}\)
\( \to {C_{M{\text{ Fe}}{\text{S}}{{\text{O}}_4}}} = {C_{M{\text{ A}}{{\text{l}}_2}{{(S{O_4})}_3}}} = \frac{{0,1}}{{0,4}} = 0,25M\)
5)
Gọi số mol \(Zn;Al\) lần lượt là \(x;y\)
\( \to 65x + 27y = 11,9{\text{ gam}}\)
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\({m_{HCl}} = 100.29,2\% = 29,2{\text{ gam }}\)
\( \to {n_{HCl}} = \frac{{29,2}}{{36,5}} = 0,8{\text{ mol = 2}}{{\text{n}}_{Zn}} + 3{n_{Al}} = 2x + 3y\)
Giải được: \(x=0,1;y=0,2\)
\( \to {m_{Zn}} = 0,1.65 = 6,5{\text{ gam}} \to {{\text{m}}_{Al}} = 5,4{\text{ gam}}\)
\({n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,4{\text{ mol}}\)
BTKL:
\({n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,4{\text{ mol}}\)
\( \to 11,9 + 100 = {m_{dd}} + 0,4.2 \to {m_{dd}} = 111,1{\text{ gam}}\)
\({n_{ZnC{l_2}}} = {n_{Zn}} = 0,1{\text{ mol;}}{{\text{n}}_{AlC{l_3}}} = {n_{Al}} = 0,2{\text{ mol}}\)
\( \to {m_{ZnC{l_2}}} = 0,1.(65 + 35,5.2) = 13,6{\text{ gam}}\)
\({m_{AlC{l_3}}} = 0,2.(27 + 35,5.3) = 26,7{\text{ gam}}\)
\( \to C{\% _{ZnC{l_2}}} = \frac{{13,6}}{{111,1}}.100\% = 12,24\% ;C{\% _{AlC{l_3}}} = \frac{{26,7}}{{111,1}}.100\% = 24,03\% \)
