Determine the total volume of the gaseous products at the end of the reaction (molar gas volume =22.4dm3 at s.t.p . mg=24 o=16 N=14

Tumia masi za atomiki zilizopewa: Mg = 24, O = 16, N = 14.
$\text{M}_r(\text{Mg(NO}_3)_2) = 24 + 2 \times (14 + 3 \times 16)$
$\text{M}_r(\text{Mg(NO}_3)_2) = 24 + 2 \times (14 + 48)$
$\text{M}_r(\text{Mg(NO}_3)_2) = 24 + 2 \times 62$
$\text{M}_r(\text{Mg(NO}_3)_2) = 24 + 124 = 148 \text{ g/mol}$

Idadi ya mola = $\frac{\text{Masi}}{\text{Masi ya mola}}$
Idadi ya mola za $\text{Mg(NO}_3)_2 = \frac{7.4 \text{ g}}{148 \text{ g/mol}} = 0.05 \text{ mol}$

Mlinganyo uliosawazishwa ni:

Kutoka mlinganyo, uwiano wa mola wa $\text{Mg(NO}_3)_2 : \text{NO}_2 : \text{O}_2$ ni $2 : 4 : 1$.
Mola za $\text{NO}_2 = \frac{4}{2} \times 0.05 \text{ mol} = 2 \times 0.05 \text{ mol} = 0.10 \text{ mol}$
Mola za $\text{O}_2 = \frac{1}{2} \times 0.05 \text{ mol} = 0.5 \times 0.05 \text{ mol} = 0.025 \text{ mol}$

Jumla ya mola za gesi = Mola za $\text{NO}_2$ + Mola za $\text{O}_2$
Jumla ya mola za gesi = $0.10 \text{ mol} + 0.025 \text{ mol} = 0.125 \text{ mol}$

Ujazo wa mola wa gesi kwenye s.t.p. ni $22.4 \text{ dm}^3/\text{mol}$.
Ujazo wa gesi = Jumla ya mola za gesi $\times$ Ujazo wa mola wa gesi
Ujazo wa gesi = $0.125 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol}$
Ujazo wa gesi = $2.8 \text{ dm}^3$