\(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{150.14,6\%}{100\%}:36,5=0,6\left(mol\right)\)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
0,1 -------> 0,2 -------->0,1---------------> 0,1
Xét \(\dfrac{0,1}{1}< \dfrac{0,6}{2}\)=> axit dư.
\(n_{HCl.dư}=0,6-0,2=0,4\left(mol\right)\)
\(m_{dd}=10+150-0,1.44=155,6\left(g\right)\)
\(C\%_{CaCl_2}=\dfrac{0,1.111.100\%}{155,6}=7,13\%\)
\(C\%_{HCl}=\dfrac{0,4.36,5.100\%}{155,6}=9,38\%\)
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