Giải thích các bước giải:

1,

\(\begin{array}{l}
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{1}{2}{n_{KMn{O_4}}} = 0,1mol \to {V_{{O_2}}} = 2,24l\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
{n_{Fe}} = 0,2mol\\
\dfrac{{{n_{Fe}}}}{3} > \dfrac{{{n_{{O_2}}}}}{2}\\
{n_{F{e_3}{O_4}}} = \dfrac{1}{2}{n_{{O_2}}} = 0,05mol \to m = 11,6g
\end{array}\)

2,

\(\begin{array}{l}
{C_2}{H_6} + \dfrac{7}{2}{O_2} \to 2C{O_2} + 3{H_2}O\\
{n_{{C_2}{H_6}}} = 0,6mol\\
{n_{{O_2}}} = \dfrac{7}{2}{n_{{C_2}{H_6}}} = 2,1mol \to {V_{{O_2}}} = 47,04l\\
{n_{C{O_2}}} = 2{n_{{C_2}{H_6}}} = 1,2mol \to {V_{C{O_2}}} = 26,88l\\
{n_{{H_2}O}} = 3{n_{{C_2}{H_6}}} = 1,8mol \to {m_{{H_2}O}} = 32,4g
\end{array}\)

1.

a. PTPU: 2KMnO4-to->K2MnO4+MnO2+O2

nKMnO4=31,65/158=0,2(mol)

Theo PT ta có: nO2=1/2nKMnO4=0,1(mol)

=>vO2=0,1x22,4=2,24(lít)

b. PTPU: 3Fe+2O2->Fe3O4

nFe=1,12/56=0,2(mol)

Theo Pt ta có: nFe3O4=1/2nO2=1/2x0.1=0.05(mol)

Xét tỷ số: 0.2/3>0.1/2=> tính theo nO2

=>mFe3O4=0.05x232=11,6(gam)

c. Ta có: nO2=0,1(mol)

=>vO2=0.1x22,4=2,24(lít)

=>vKK=5vO2=5x2,24=11,2(lít)

2. PTPU: 2C₂H₆+7O₂→6H₂O+4CO₂

nC2H6=13,44/22,4=0,6(mol)

Theo PT ta có:

nO2=7/2nC2H6=7/2x0,6=2,1(mol)

nCO2=2nC2H6=0,6x2=1,2(mol)

nH2O=3nC2H6=1,8(mol)

=>vO2=2,1x22,4=47,04(lít)

=>vCO2=1,2x22,4=26,88(lít)

=>mH2O=1,8x18=32,4(gam)