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Ta có: \(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{400.3,65\%}{36,5}=0,4\left(mol\right)\)
PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}=\dfrac{0,4}{2}\), ta được pư vừa đủ.
Theo PT: \(n_{CaCl_2}=n_{CO_2}=n_{CaCO_3}=0,2\left(mol\right)\)
Có: m dd sau pư = 20 + 400 - 0,2.44 = 411,2 (g)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{0,2.111}{411,2}.100\%\approx5,4\%\)
Ta có: \(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)
PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
______0,2_______0,4_____0,2____0,2 (mol)
a, \(m_{ddHCl}=\dfrac{0,4.36,5}{14,6\%}=100\left(g\right)\)
b, \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
c, m dd sau pư = 20 + 100 - 0,2.44 = 111,2 (g)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{0,2.111}{111,2}.100\%\approx19,96\%\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
c, \(C_{M_{ZnCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
b, Ta có: 80nCuO + 81nZnO = 12,1 (1)
Theo PT: \(n_{HCl}=2n_{CuO}+2n_{ZnO}=0,2.1,5=0,3\left(mol\right)\left(2\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{CuO}=0,05\left(mol\right)\\n_{ZnO}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{12,1}.100\%\approx33,06\%\\\%m_{ZnO}\approx66,94\%\end{matrix}\right.\)