Đáp án:
\(b,\ C\%_{CaCl_2}=9,27\%\\ C\%_{HCl}(\text{dư})=1,53\%\)
Giải thích các bước giải:
\(a,\ PTHH:CaCO_3+2HCl\to CaCl_2+CO_2+H_2O\\ b,\ n_{CaCO_3}=\dfrac{10}{100}=0,1\ mol.\\ n_{HCl}=\dfrac{114,1\times 8\%}{36,5}=0,25\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,1}{1}<\dfrac{0,25}{2}\\ ⇒HCl\ dư.\\ n_{HCl}(\text{dư})=0,25-(0,1\times 2)=0,05\ mol.\\ Theo\ pt:\ n_{CaCl_2}=n_{CO_2}=n_{CaCO_3}=0,1\ mol.\\ m_{\text{dd spư}}=m_{CaCO_3}+m_{HCl}-m_{CO_2}=10+114,1-0,1\times 44=119,7\ g.\\ ⇒C\%_{CaCl_2}=\dfrac{0,1\times 111}{119,7}\times 100\%=9,27\%\\ C\%_{HCl}(\text{dư})=\dfrac{0,05\times 36,5}{119,7}\times 100\%=1,53\%\)
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