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Gọi x là số mol \(SO_2\), y là số mol \(SO_3\)
có: 64x + 80y = 1,408 (1)
2,2 g khí X = 1 g khí O\(_2\) có thể tích bằng nhau
=> 1,408g X = 0,64 g khí \(O_2\)
<=> x + y = \(\dfrac{0,64}{32}=0,02\) (2)
Từ (1), (2) suy ra:
x = 0,012
y = 0,008
=> \(n_{O\left(trong.hh.X\right)}=0,012.2+0,008.3=0,048\left(mol\right)\)
=> \(n_{Fe_2\left(SO_4\right)_3}=0,048:12=0,004\left(mol\right)\)
=> \(m_{Fe_2\left(SO_4\right)_3}=0,004.400=1,6\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe_3O_4}=\dfrac{23.2}{232}=0.1\left(mol\right)\)
\(PTHH:Fe_3O_4+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+FeSO_4+4H_2O\)
\(...........0.1....................0.1.......0.1.........0.1\)
\(m_{FeSO_4}=0.1\cdot152=15.2\left(g\right)\)
\(m_{Fe_2\left(SO_4\right)_3}=0.1\cdot400=40\left(g\right)\)
\(m_{H_2O}=0.1\cdot18=1.8\left(g\right)\)
a)
$Fe_3O_4 + 4H_2SO_4 \to FeSO_4 + Fe_2(SO_4)_3 + 4H_2O$
b)
Theo PTHH :
$n_{FeSO_4} = n_{Fe_2(SO_4)_3} = n_{Fe_3O_4} = \dfrac{23,2}{232} = 0,1(mol)$
$m_{FeSO_4} = 0,1.152 = 15,2(gam)$
c)
$m_{Fe_2(SO_4)_3} = 0,1.400 = 40(gam)$
d)
$n_{H_2O} = 4n_{Fe_3O_4} = 0,4(mol)$
$m_{H_2O} = 0,4.18 = 7,2(gam)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Giả sử có 100 gam hỗn hợp
=> \(m_S=\dfrac{100.22,61}{100}=22,61\left(g\right)\)
=> \(n_S=\dfrac{22,61}{32}=\dfrac{2261}{3200}\left(mol\right)\)
Mà nO = 4nS
=> \(n_O=\dfrac{2261}{800}\left(mol\right)\)
\(\%m_O=\dfrac{\dfrac{2261}{800}.16}{100}.100\%=45,22\%\)
b)
\(n_{Fe}=\dfrac{18.10^{24}}{6.10^{23}}=30\left(mol\right)\)
=> \(n_{Fe_2\left(SO_4\right)_3}=15\left(mol\right)\)
Gọi số mol CuSO4 là x (mol)
=> mhh = 160x + 6000 (g)
nS = 15.3 + x = x + 45 (mol)
\(\%m_S=\dfrac{\left(x+45\right).32}{160x+6000}.100\%=22,61\%\)
=> x = 20 (mol)
mhh = 160.20 + 6000 = 9200 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: nC= \(\dfrac{1,9926.10^{-23}}{12}\) = 1,6605.10-24
=> mNa = ( 1,6605. 10-24) .23= 3,81915.10-23
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M_{Fe_2\left(SO_4\right)_3}=400\left(g\text{/}mol\right)\)
\(\%Fe=\dfrac{56\cdot2}{400}\cdot100\%=28\%\)
\(\%S=\dfrac{32\cdot3}{400}\cdot100\%=24\%\)
\(\%O=100-28-24=48\%\)