a, Ta có: \(n_{CaCO_3}=\dfrac{300}{100}=3\left(mol\right)\)
\(m_{HCl}=400.7,3\%=29,2\left(g\right)\Rightarrow n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)
PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
Xét tỉ lệ: \(\dfrac{3}{1}>\dfrac{0,8}{2}\), ta được CaCO3 dư.
Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{HCl}=0,4\left(mol\right)\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)
b, Theo PT: \(n_{CaCO_3\left(pư\right)}=n_{CaCl_2}=\dfrac{1}{2}n_{HCl}=0,4\left(mol\right)\)
Ta có: m dd sau pư = 0,4.100 + 400 - 0,4.44 = 422,4 (g)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{0,4.111}{422,4}.100\%\approx10,51\%\)
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