Đáp án:

$a)V_{dd_{KOH}}\approx 57,42(ml)$

$b)C\%_{K_2SO_4}=6,525\%$

Giải thích các bước giải:

$a)PTHH:2KOH+H_2SO_4\to K_2SO_4+2H_2O$

$m_{H_2SO_4}=200.14,7\%=29,4(g)$

$\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3(mol)$

Theo PT: $n_{KOH}=2n_{H_2SO_4}=0,6(mol)$

$\Rightarrow m_{KOH}=0,6.56=33,6(g)$

$\Rightarrow m_{dd_{KOH}}=\dfrac{m_{KOH}}{C\%}=\dfrac{33,6}{5,6\%}=600(g)$

$\Rightarrow V_{dd_{KOH}}=\dfrac{V_{dd}}{D}=\dfrac{600}{10,45}\approx 57,42(ml)$

$b)$ Theo PT: $n_{K_2SO_4}=n_{H_2SO_4}=0,3(mol)$

$\Rightarrow m_{K_2SO_4}=0,3.174=52,2(g)$

Ta có $m_{\text{dd sau p/ứ}}=m_{dd_{KOH}}+m_{dd_{H_2SO_4}}$

$\Rightarrow m_{\text{dd sau p/ứ}}=600+200=800(g)$

$\Rightarrow C\%_{K_2SO_4}=\dfrac{52,2}{800}.100\%=6,525\%$

Bạn tham khảo!
Đáp án:

$a)57,42ml$

$b)6,525$ $\text{%}$

Giải thích các bước giải:

 Ta có PTHH sau:

$2KOH+H_2SO_4$ $\rightarrow$ $K_2SO_4+2H_2O$

$+)$ Dung dịch muối: $K_2SO_4$

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Có $m_{H_2SO_4}$ $=$ $\dfrac{200.14,7}{100}$ $=29,4g$

Vậy $n_{H_2SO_4}$ $=$ $\dfrac{29,4}{98}$ $=0,3$ mol

Dựa vào PTHH) $n_{KOH}$ $=0,3.2=0,6$ mol

a) Vậy $m_{KOH}$ $=0,6.56=33,6g$

$\rightarrow$ $mdd_{KOH}$ $=$ $\dfrac{33,6.100}{5,6}$ $=600g$

Vậy $V_{KOH}$ $=$ $\dfrac{m}{D}$ $=$ $\dfrac{600}{10,45}$ $=57,42ml$

b) Dựa vào PTHH) $n_{K_2SO_4}$ $=0,3$ mol

Vậy $m_{K_2SO_4}$ $=0,3.174=52,2g$

Lại có $mdd_{\text{sau}}$ $=$ $mdd_{KOH}$ $+$ $mdd_{H_2SO_4}$ $=600+200=800g$

$\rightarrow$ $\text{C%}$ $K_2SO_4$ $=$ $\dfrac{52,2}{800}$ $.100=6,525$ $\text{%}$