13)

Phản ứng xảy ra:

\(KOH + HCl\xrightarrow{{}}KCl + {H_2}O\)

Ta có:

\({n_{HCl}} = 0,25.1,5 = 0,375{\text{ mol = }}{{\text{n}}_{KOH}} = {n_{KCl}}\)

\( \to {V_{dd{\text{ KOH}}}} = \frac{{0,375}}{2} = 0,1875{\text{ lít}}\)

\({V_{dd{\text{ sau phản ứng}}}} = 0,1875 + 0,25 = 0,4375{\text{ lít}}\)

\( \to {C_{M{\text{ KCl}}}} = \frac{{0,375}}{{0,4375}} = 0,857M\)

14)

Phản ứng xảy ra:

\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

\(Cu\) không phản ứng với axit loãng.

\( \to {n_{{H_2}}} = \frac{3}{2}{n_{Al}}  = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\)

\( \to {n_{Al}} = 0,2{\text{ mol}}\)

\( \to {m_{Al}} = 0,2.27 = 5,4{\text{ gam}}\)

\( \to {m_{Cu}} = 10 - {m_{Al}} = 10 - 5,4 = 4,6{\text{ gam}}\)

\({n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}}} = 0,3.98 = 29,4{\text{ gam}}\)

\( \to {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{29,4}}{{20\% }} = 147{\text{ gam}}\)

Đáp án:

Bài 13: 

\(\begin{array}{l} a,\ V_{\text{dd KOH}}=0,1875\ lít=187,5\ ml.\\ b,\ C_{M_{KCl}}=\dfrac{6}{7}\ M.\end{array}\)

Bài 14:

\(\begin{array}{l} b,\ m_{Al}=5,4\ g.\\ m_{Cu}=4,6\ g.\\ c,\ m_{\text{dd H$_2$SO$_4$}}=147\ g.\end{array}\)

Giải thích các bước giải:

Bài 13:

\(\begin{array}{l} a,\ PTHH:KOH+HCl\to KCl+H_2O\\ n_{HCl}=0,25\times 1,5=0,375\ mol.\\ Theo\ pt:\ n_{KOH}=n_{HCl}=0,375\ mol.\\ \Rightarrow V_{\text{dd KOH}}=\dfrac{0,375}{2}=0,1875\ lít=187,5\ ml.\\ b,\ Theo\ pt:\ n_{KCl}=n_{HCl}=0,375\ mol.\\ V_{\text{dd spư}}=V_{KOH}+V_{HCl}=187,5+250=437,5\ ml=0,4375\ lít.\\ \Rightarrow C_{M_{KCl}}=\dfrac{0,375}{0,4375}=\dfrac{6}{7}\ M.\end{array}\)

Bài 14:

\(\begin{array}{l} a,\ PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑\\ b,\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\ mol.\\ Theo\ pt:\ n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\ mol.\\ \Rightarrow m_{Al}=0,2\times 27=5,4\ g.\\ \Rightarrow m_{Cu}=10-5,4=4,6\ g.\\ c,\ Theo\ pt:\ n_{H_2SO_4}=n_{H_2}=0,3\ mol.\\ \Rightarrow m_{H_2SO_4}=0,3\times 98=29,4\ g.\\ \Rightarrow m_{\text{dd H$_2$SO$_4$}}=\dfrac{29,4}{20\%}=147\ g.\end{array}\)

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