1. nS=24/32=0,75(m0l)

S+O2-to->SO2

nO2=nS=0,75(mol)

=>vO2=0,75x22,4=18,8(lít)

b. nSO2=nS=0,75(mol)

=>mSO2=0,75x64=48(gam)

2.

a. nO2=4,48/22,4=0,2(mol)

2Cu+O2-to->2CuO

nCu=2nO2=0,2x2=0,4(mol)

=>mCu=0,4x64=25,6(gam)

b. nCuO=nCu=0,4(mol)

=>CuO=0,4x80=32(gam)

3. 

a. nFe=33,6/56=0,6(mol)

3Fe+2O2-to->Fe3O4

nO2=2/3nFe=2/3x0,6=0,4(mol)

=>vO2=0,4x22,4=8,96(lít)

b. nFe3O4=1/3nFe=1/3x0,6=0,2(mol)

=>mFe3O4=0,2x232=46,4(gam)

Đáp án:

1,

\(\begin{array}{l}
S + {O_2} \to S{O_2}\\
{n_S} = 0,75mol = {n_{{O_2}}}\\
{V_{{O_2}}} = 16,8l\\
{n_{S{O_2}}} = {n_S} = 0,75mol \to {m_{S{O_2}}} = 48g
\end{array}\)

2,

\(\begin{array}{l}
Cu + \frac{1}{2}{O_2} \to CuO\\
{n_{{O_2}}} = 0,2mol \to {n_{Cu}} = 2{n_{{O_2}}} = 0,4mol \to {m_{Cu}} = 25,6g\\
{n_{CuO}} = {n_{Cu}} = 0,4mol \to {m_{CuO}} = 32g
\end{array}\)

3,

\(\begin{array}{l}
3Fe + 2{O_2} \to F{e_3}{O_4}\\
{n_{Fe}} = 0,6mol \to {n_{{O_2}}} = \frac{2}{3}{n_{Fe}} = 0,4mol \to V = 8,96l\\
{n_{F{e_3}{O_4}}} = \frac{1}{3}{n_{Fe}} = 0,2mol \to {m_{F{e_3}{O_4}}} = 46,4g
\end{array}\)