\(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
PTHH: CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
0,1--->0,2------>0,1---->0,1
=> mdd HCl = \(\dfrac{0,2.36,5}{3,65\%}=200\left(g\right)\)
mdd sau pư = 10 + 200 - 0,1.44 = 205,6 (g)
mCaCl2 = 0,1.111 = 11,1 (g)
=> \(C\%_{CaCl_2}=\dfrac{11,1}{205,6}.100\%=5,4\%\)
\(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
PTHH: CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O
0,1------->0,2--------->0,1------>0,1
\(\rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{3,65\%}=200\left(g\right)\\ \rightarrow m_{dd}=200+10-0,1.44=205,4\left(g\right)\\ \rightarrow C\%_{CaCl_2}=\dfrac{0,1.111}{205,4}.100\%=5,4\%\)
