CaCO3--t0---> CaO + CO2
MgCO3--t0---> MgO + CO2
Gọi nCaCO3= a mol
nMgCO3= b mol
=> 100a+ 84b=7,44
Ta có a+ b=nCO2= 1,792/22,4=0,08 mol
=> a=0,045 mol ; b=0,035 mol
=> %mCaCO3=0,045.100.100/7,44=60,48%
=> %mMgCO3 = 100- 60,48=39,52%
nCO2 = \(\dfrac{1,792}{22,4}\) = 0,08 mol
CaCO3 \(^{to}\rightarrow\) CaO + CO2
x----------------------->x
MgCO3 \(^{to}\rightarrow\) MgO + CO2
y---------------------->y
ta có : \(\left\{{}\begin{matrix}100x+84y=7,44\\x+y=0,08\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,045mol\\y=0,035mol\end{matrix}\right.\)
%CaCO3 = \(\dfrac{0,045.100}{7,44}.100\%\) = 60,48%
=>%MgCO3 = 100 % - 60,48% 0 39,52%
CaCO3 --to--➢ CaO + CO2 (1)
MgCO3 --to--➢ MgO + CO2 (2)
\(n_{CO_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
Gọi \(x,y\) lần lượt là số mol của CaCO3 và MgCO3
Ta có: \(\left\{{}\begin{matrix}100x+84y=7,44\\x+y=0,08\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,045\\y=0,035\end{matrix}\right.\)
Vậy \(n_{CaCO_3}=0,045\left(mol\right)\) ; \(n_{MgCO_3}=0,035\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,045\times100=4,5\left(g\right)\)
\(\Rightarrow\%CaCO_3=\dfrac{4,5}{7,44}\times100\%=60,48\%\)
\(m_{MgCO_3}=0,035\times84=2,94\left(g\right)\)
\(\Rightarrow\%MgCO_3=\dfrac{2,94}{7,44}\times100\%=39,52\%\)
