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![](https://rs.olm.vn/images/avt/0.png?1311)
Vì Mg có tính khử rất mạnh nên trong Y có thể có muối (NH4)2SO4
BTNT N: 2nFe(NO3)2 = nNO + nNH4 => 2y = 0,02 + nNH4 => nNH4 = 2y – 0,02
BTNT H: 2nH2SO4 = 2nH2 + 4nNH4 + 2nH2O => 2 . 0,39 = 2 . 0,05 + 4 (2y – 0,02) + 2nH2O
=> nH2O = 0,38 – 4y
BTNT O: nFeO + 6nFe(NO3)2 = nNO + nH2O
=> x + 6y = 0,02 + 0,38 – 4y => x + 10y = 0,4
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)
c, \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Mg}+n_{MgO}=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6\%}=75\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\left(1\right)\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\left(2\right)\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT (1): \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=m_{hh}-m_{Mg}=4,4-2,4=2\left(g\right)\)
Bạn tham khảo nhé!
mFeCO3= 9/38 x 15,2= 3,6(g) -> nFeCO3= 3,6/116= 14/29(mol)
=> mMg= 15,2- 3,6= 11,6(g) -> nMg=11,6/24= 29/60(mol)
PTHH: FeCO3 + H2SO4 -> FeSO4 + CO2 + H2O
14/29____________14/29_____________14/29(mol)
Mg + H2SO4 -> MgSO4 + H2
29/60____29/60____________29/60(mol)
Ta có:
\(M_Y=\dfrac{\dfrac{14}{29}.44+\dfrac{29}{60}.2}{\dfrac{14}{29}+\dfrac{29}{60}}=22,988\left(\dfrac{g}{mol}\right)\\ \rightarrow d_{\dfrac{Y}{H2}}=\dfrac{22,988}{2}=11,494\)
nH2SO4(p.ứ)=14/29 + 29/60=0,966(mol)
Gọi x,y lần lượt là số mol FeCO3 và Mg
Ta có : \(\left\{{}\begin{matrix}116x+24y=15,2\\\dfrac{116x}{24y}=\dfrac{9}{29}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\dfrac{9}{290}\\y=\dfrac{29}{60}\end{matrix}\right.\)
\(FeCO_3+H_2SO_4\rightarrow FeSO_4+CO_2+H_2O\)
\(\dfrac{9}{290}\)------>\(\dfrac{9}{290}\)--------------------->\(\dfrac{9}{290}\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(\dfrac{29}{60}\)----->\(\dfrac{29}{60}\)-------------------->\(\dfrac{29}{60}\)
=> \(M_Y=\dfrac{\dfrac{9}{290}.44+\dfrac{29}{60}.2}{\dfrac{9}{290}+\dfrac{29}{60}}=4,53\)
=>dY/H2= \(\dfrac{4,53}{2}=2,265\)
\(n_{H_2SO_4}=\dfrac{9}{290}+\dfrac{29}{60}=\dfrac{179}{348}\left(mol\right)\)