a, Gọi \(\left\{{}\begin{matrix}n_{MgCO_3}=a\left(mol\right)\\n_{CaCO_3}=b\left(mol\right)\end{matrix}\right.\)
\(n_{HCl}=1,2.0,5=0,6\left(mol\right)\)
PTHH:
MgCO3 + 2HCl ---> MgCl2 + CO2 + H2O
a----------->2a--------->a------->a
CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O
b----------->2b-------->a-------->b
Hệ pt \(\left\{{}\begin{matrix}84a+100b=28,4\\2a+2b=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
=> VCO2 = (0,1 + 0,2).22,4 = 6,72 (l)
b, \(\left\{{}\begin{matrix}m_{MgCO_3}=0,1.84=8,4\left(g\right)\\m_{CaCO_3}=0,2.100=20\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{8,4}{28,4}.100\%=29,58\%\\\%m_{CaCO_3}=100\%-29,58\%=70,42\%\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}C_{M\left(MgCl_2\right)}=\dfrac{0,1}{0,5}=0,2M\\C_{M\left(CaCl_2\right)}=\dfrac{0,2}{0,5}=0,4M\end{matrix}\right.\)
\(n_{HCl}=0,5.1,2=0,6\left(mol\right)\)
gọi \(n_{MgCO_3}:a,n_{CaCO_3}:b\left(a,b>0\right)\)
=> 84a + 100b = 28,4(g)
pthh: \(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)
a 2a
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
b 2b
=> 2a + 2b = 0,6
=> \(\left\{{}\begin{matrix}84a+100b=28,4\\2a+2b=0,6\end{matrix}\right.\)
=> a= 0,1 , b = 0,2
=> \(m_{MgCO_3}=\dfrac{0,1.84}{28,4}.100\%=29,577\%\\
m_{CaCO_3}=100\%-29,577\%=70,423\)
theo 2 pt trên => nCO2 = \(\dfrac{1}{2}n_{HCl}\) = 0,3 (mol)
=> \(V_{CO_2}=0,3.22,4=6,72\left(l\right)\)