Question

Giúp bài 4

Answer 1

\(n_{CO_2}=\dfrac{88000}{44}=2000\left(mol\right)\\ CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\\ MgCO_3\rightarrow\left(t^o\right)MgO+CO_2\\ Đặt:n_{CaCO_3}=a\left(mol\right);n_{MgCO_3}=b\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}56a+40b=104000\\a+b=2000\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1500\\b=500\end{matrix}\right.\\ \Rightarrow m_{đá}=m_{CaCO_3}+m_{MgCO_3}=1500.100+84.500=192000\left(g\right)=192\left(kg\right)\)