Đáp án:

a. \(m_{{CaCO_3}}=150 \ \text{gam}\)

b. \(\text{C%}_{CaCl_2}=19,1\%\)

Giải thích các bước giải:

 4.

a. \(CaCO_3+2HCl\to CaCl_2+CO_2+H_2O\)

\(n_{CO_2}=\frac{33,6}{22,4}=1,5\ \text{mol}\to n_{HCl}=2n_{CO_2}=3\ \text{mol}; n_{CaCO_3}=1,5\ \text{mol}\)

\(\to m_{CaCO_3}=1,5\cdot 100=150\ \text{gam}\)

\(\to m_{\text{đá vôi}}=\frac{150}{80\%}=187,5\ \text{gam}\)

b. \(n_{HCl}=2n_{CO_2}=3\ \text{mol}\to m_{\text{dung dịch HCl}}=\frac{3\cdot 36,5}{14,6\%}=750\ \text{gam}\)

BTKL \(\to m_{\text{dung dịch sau phản ứng}}=187,5+750-1,5\cdot 44=871,5\ \text{gam}\)

\(n_{CaCl_2}=n_{CO_2}=1,5\ \text{mol}\to m_{CaCl_2}=1,5\cdot 111=166,5\ \text{gam}\)

\(\to \text C\%_{CaCl_2}=\frac{166,5}{871,5}\cdot 100\%=19,1\%\)