Đáp án:
\(\begin{array}{l} b,\ m_{CuO}=16\ g.\\ c,\ C_{M_{Ba(OH)_2\text{(dư)}}}=\dfrac{1}{9}\ M.\\ C_{M_{BaCl_2}}=\dfrac{4}{9}\ M.\end{array}\)
Giải thích các bước giải:
Câu 4:
\(\begin{array}{l} a,\\ PTHH:\\ CuCl_2+Ba(OH)_2\to Cu(OH)_2\downarrow+BaCl_2\ (1)\\ Cu(OH)_2\xrightarrow{t^o} CuO+H_2O\ (2)\\ b,\\ n_{CuCl_2}=0,2\times 1=0,2\ mol.\\ n_{Ba(OH)_2}=0,25\times 1=0,25\ mol.\\ \text{Lập tỉ lệ}\ n_{CuCl_2}:n_{Ba(OH)_2}=\dfrac{0,2}{1}<\dfrac{0,25}{1}\\ \Rightarrow \text{Ba(OH)$_2$ dư.}\\ Theo\ pt:\ n_{CuO}=n_{Cu(OH)_2}=n_{CuCl_2}=0,2\ mol.\\ \Rightarrow m_{CuO}=0,2\times 80=16\ g.\\ c,\\ n_{Ba(OH)_2\text{(dư)}}=0,25-0,2=0,05\ mol.\\ Theo\ pt\ (1):\ n_{BaCl_2}=n_{CuCl_2}=0,2\ mol.\\ V_{\text{dd spư}}=V_{\text{dd CuCl$_2$}}+V_{\text{dd Ba(OH)$_2$}}=200+250=450\ ml=0,45\ lít.\\ \Rightarrow C_{M_{Ba(OH)_2\text{(dư)}}}=\dfrac{0,05}{0,45}=\dfrac{1}{9}\ M.\\ C_{M_{BaCl_2}}=\dfrac{0,2}{0,45}=\dfrac{4}{9}\ M.\end{array}\)
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