a) Gọi \(\left\{{}\begin{matrix}n_{MgCO_3}=x\left(mol\right)\\n_{CaCO_3}=y\left(mol\right)\end{matrix}\right.\Rightarrow84x+100y=63,6\left(1\right)\)
Ta có: \(n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
PTHH: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2\uparrow+H_2O\)
x-------->2x--------------------->x
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\)
y-------->2y--------------------->y
`=> x + y = 0,7 (2)`
Từ \(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,3\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,4.84}{63,6}.100\%=52,83\%\\\%m_{CaCO_3}=100\%-52,83\%=47,17\%\end{matrix}\right.\)
b) \(V_{ddHCl}=\dfrac{2.\left(0,3+0,4\right)}{4}=0,35\left(l\right)\)
