goi a la khoi luong cua CaCO3
b la khoi luong cua MgCO3
\(n_{CaCO_3}=\dfrac{a}{100}\left(mol\right)\)
\(n_{MgCO_3}=\dfrac{b}{84}\left(mol\right)\)
\(CaCO_3\rightarrow CaO+CO_2\) (1)
de: \(\dfrac{a}{100}\rightarrow\dfrac{a}{100}\rightarrow\dfrac{a}{100}\left(mol\right)\)
\(MgCO_3\rightarrow MgO+CO_2\) (2)
de: \(\dfrac{b}{84}\rightarrow\dfrac{b}{84}\rightarrow\dfrac{b}{84}\left(mol\right)\)
theo de: \(m_{CaO}+m_{MgO}=\dfrac{a+b}{2}\)
\(\Rightarrow0,56a+\dfrac{10}{21}b=\dfrac{a+b}{2}\) (3)
theo ĐLBTKL: \(m_{CO_2}=m_{CO_2\left(1\right)}+m_{CO_2\left(2\right)}=\dfrac{a+b}{2}\)
\(\Leftrightarrow0,44a+\dfrac{11}{21}b=\dfrac{a+b}{2}\) (4)
tu (3) va (4) \(\Rightarrow0,56a+\dfrac{10}{21}b=0,44a+\dfrac{11}{21}b\)
\(\Leftrightarrow0,12a=\dfrac{1}{21}b\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{1}{21}:0,12=\dfrac{25}{63}\)
\(\Rightarrow m_{CaCO_3}=25\) g va \(m_{MgCO_3}=63\) g
\(\%m_{CaCO_3}=\dfrac{25}{25+63}.100\approx28,4\%\)
\(\%m_{MgCO_3}=100\%-\%m_{CaCO_3}=100-28,4=71,6\%\)
minh lm hoi tat nen ban thong cam
