Hỗn hợp X: \(\left\{{}\begin{matrix}CaCO_3:a\left(mol\right)\\FeCO_3:b\left(mol\right)\end{matrix}\right.\)
\(CaCO_3\left(a\right)+H_2SO_4\left(a\right)\rightarrow CaSO_4\left(a\right)+CO_2\left(a\right)+H_2O\)
\(FeCO_3\left(b\right)+H_2SO_4\left(b\right)\rightarrow FeSO_4\left(b\right)+CO_2\left(b\right)+H_2O\)
\(n_{CO_2}=0,03\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}100a+116b=3,32\\a+b=0,03\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=0,01\left(mol\right)\\b=0,02\left(mol\right)\end{matrix}\right.\)
Suy ra phần trăm về khối lượng của mỗi muối trong hh đầu
Theo PTHH: \(n_{H_2SO_4}=n_{CO_2}=0,03\left(mol\right)\)
\(\Rightarrow m_{H_2SO_3}=0,03.98=2,94\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{2,94}{100}.100=2,94\%\)
\(mddsau=3,32+100-0,03.44=102(g)\)
Theo PTHH: \(\left\{{}\begin{matrix}n_{CaSO_4}=a=0,01\left(mol\right)\\n_{FeSO_4}=b=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{CaSO_4}=\dfrac{0,01.136}{102}.100=1,33\%\\C\%_{FeSO_4}=\dfrac{0,02.152}{102}.100=2,98\%\end{matrix}\right.\)
\(n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
Gọi x, y lần lượt là số mol của CaCO3 , FeCO3
Pt: \(CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2\uparrow+H_2O\)(1)
x -------------> 0,01 ------> 0,01-----> x
\(FeCO_3+H_2SO_4\rightarrow FeSO_4+CO_2+H_2O\) (2)
y -----------> 0,02 -----------> 0,02 ------> y
\(\left(1\right)\left(2\right)\Rightarrow\left\{{}\begin{matrix}100x+116y=3,32\\x+y=0,03\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,02\end{matrix}\right.\)
\(\%CaCO_3=\dfrac{0,01.100}{3,32}.100=30,12\%\)
\(\Rightarrow\%FeCO_3=100-30,12=69,88\%\)
\(\Sigma_{m_{dd\left(spu\right)}}=3,32+100-0,03.44=102\left(g\right)\)
\(C\%_{H_2SO_4\left(spu\right)}=\dfrac{0,03.98.100}{102}=2,88\%\)
\(C\%_{CaSO_4}=\dfrac{0,01.136.100}{102}=1,33\%\)
\(C\%_{FeSO_4}=\dfrac{0,02.152.100}{102}=2,98\%\)
