a. PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
\(n_{CO_2}=0,1\left(mol\right)\Rightarrow n_{HCl}=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=7,3\left(g\right)\) \(\Rightarrow C\%=\dfrac{7,3.100}{200}=3,65\%\)
b. \(n_{CO_2}=0,1\left(mol\right)\Rightarrow n_{CaCO_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=100.0,1=10\left(g\right)\)
c. \(n_{CaCl_2}=0,1\left(mol\right)\Rightarrow m_{CaCl_2}=111.0,1=11,1\left(g\right)\)
\(C\%=11,1.\dfrac{100}{200}=5,55\%\)
a, Ta có: \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
_____0,1______0,2_____0,1___0,1 (mol)
\(\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{7,3}{200}.100\%=3,65\%\)
b, \(m_{CaCO_3}=0,1.100=10\left(g\right)\)
c, Ta có: m dd sau pư = mCaCO3 + m dd HCl - mCO2 = 10 + 200 - 0,1.44 = 205,6(g)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{0,1.111}{205,6}.100\%\approx5,4\%\)
