Question

Answer 1

a)

\(n_{FeS}=\dfrac{3,52}{88}=0,04\left(mol\right)\)

\(n_{O_2}=\dfrac{22,4.20\%}{22,4}=0,2\left(mol\right)\)

PTHH: 4FeS + 7O₂ --t^o--> 2Fe₂O₃ + 4SO₂

Xét tỉ lệ: \(\dfrac{0,04}{4}< \dfrac{0,2}{7}\) => FeS hết, O₂ dư

PTHH: 4FeS + 7O₂ --t^o--> 2Fe₂O₃ + 4SO₂

            0,04-->0,07------->0,02---->0,04

\(m_A=m_{Fe_2O_3}=0,02.160=3,2\left(g\right)\)

b)\(n_{N_2}=\dfrac{22,4}{22,4}-0,2=0,8\left(mol\right)\)

B gồm \(\left\{{}\begin{matrix}N_2:0,8\left(mol\right)\\SO_2:0,04\left(mol\right)\\O_{2\left(dư\right)}:0,13\left(mol\right)\end{matrix}\right.\)

\(M_B=\dfrac{0,8.28+0,04.64+0,13.32}{0,8+0,04+0,13}=\dfrac{2912}{97}\left(g/mol\right)\)=> \(d_{B/H_2}=\dfrac{\dfrac{2912}{97}}{2}=\dfrac{1456}{97}\)

Answer 2

Answer 3

`a)`

\(n_{FeS}=\dfrac{3,52}{88}=0,04\left(mol\right)\)

Trong 22,4 lít không khí có: \(\left\{{}\begin{matrix}V_{O_2}=\dfrac{20}{100}.22,4=4,48\left(l\right)\\V_{N_2}=22,4-4,48=17,92\left(l\right)\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{N_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\end{matrix}\right.\)

PTHH: 

                      \(4FeS+7O_2\xrightarrow[]{t^o}2Fe_2O_3+4SO_2\uparrow\)

ban đầu         0,04      0,2

sau pư            0         0,13       0,02           0,04

`=> m_A = m_{Fe_2O_3} = 0,02.160 = 3,2 (g)`

`b)`

Trong khí B có \(\left\{{}\begin{matrix}SO_2:0,04mol\\O_2:0,13mol\\N_2:0,8\left(mol\right)\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}n_B=0,04+0,13+0,8=0,97\left(mol\right)\\m_B=0,04.64+0,13.32+0,8.28=29,12\left(g\right)\end{matrix}\right.\)

$\Rightarrow M_B = \dfrac{29,12}{0,97} = 30,02 (g/mol)$

$\Rightarrow d_{B/H_2} = \dfrac{30,02}{2} = 15,01$