Câu 2:
\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:\)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
0,2<------0,4<----0,2<---------------0,2
a. \(m=m_{CaCO_3}=0,2.100=20\left(g\right)\)
b. \(m_{dd.HCl}=\dfrac{0,4.36,5.100\%}{10\%}=146\left(g\right)\)
c. \(C\%_{CaCl_2}=\dfrac{0,2.111.100\%}{20+146-0,2.44}=14,12\%\)
`HaNa♬D`
Câu 3 :
Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
a) Theo Pt : \(n_{H2}=n_{Mg}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\%m_{Mg}=\dfrac{0,3.24}{14,4}.100\%=50\%\)
\(\%m_{FeO}=100\%-50\%=50\%\)
b) Pt : \(3H_2+Fe_2O_3\xrightarrow[]{t^o}2Fe+3H_2O\)
Theo Pt : \(n_{Fe2O3\left(tt\right)}=\dfrac{1}{3}n_{H2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)
\(\Rightarrow H\%=\dfrac{0,1.56}{8}.100\%=70\%\)
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