Đáp án:
\(\begin{array}{l} b,\ m_{\text{dd HCl}}=400\ g.\\ c,\ C\%_{BaCl_2}=18,04\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\\ PTHH:BaCO_3+2HCl\to BaCl_2+CO_2↑+H_2O\\ b,\\ n_{BaCO_3}=\dfrac{78,8}{197}=0,4\ mol.\\ Theo\ pt:\ n_{HCl}=2n_{BaCO_3}=0,8\ mol.\\ \Rightarrow m_{\text{dd HCl}}=\dfrac{0,8\times 36,5}{7,3\%}=400\ g.\\ c,\\ Theo\ pt:\ n_{CO_2}=n_{BaCO_3}=0,4\ mol.\\ \Rightarrow m_{\text{dd spư}}=m_{BaCO_3}+m_{\text{dd HCl}}-m_{CO_2}\\ \Rightarrow m_{\text{dd spư}}=78,8+400-0,4\times 44=461,2\ g.\\ Theo\ pt:\ n_{BaCl_2}=n_{BaCO_3}=0,4\ mol.\\ \Rightarrow C\%_{BaCl_2}=\dfrac{0,4\times 208}{461,2}\times 100\%=18,04\%\end{array}\)
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