CaCO3 + 2HCl → CaCl2 + CO2 + H2O
\(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
\(m_{HCl}=100\times7,3\%=7,3\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
Theo PT: \(n_{CaCO_3}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{CaCO_3}=\dfrac{1}{2}n_{HCl}\)
Vì \(\dfrac{1}{2}=\dfrac{1}{2}\) ⇒ CaCO3 và HCl đều phản ứng hết
Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,1\times22,4=2,24\left(l\right)\)
\(m_{CO_2}=0,1\times44=4,4\left(g\right)\)
Theo PT: \(n_{CaCl_2}=n_{CaCO_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{CaCl_2}=0,1\times111=11,1\left(g\right)\)
\(m_{dd}saupư=m_{CaCO_3}+m_{ddHCl}-m_{CO_2}=10+100-4,4=105,6\left(g\right)\)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{11,1}{105,6}\times100\%=10,51\%\)
