Given in the question that molar mass of Na₂CO₃ = 106 g

$$ \therefore $$ Molarity of solution = $${{25.3 \times 1000} \over {106 \times 250}}$$

= 0.9547 M = 0.955 M

Na₂CO₃ $$ \to $$ 2Na⁺ + $$CO_3^{2-}$$

[Na⁺] = 2[Na₂CO₃] = 2 $$ \times $$ 0.955 = 1.910 M

[$$CO_3^{2-}$$] = [Na₂CO₃] = 0.955 M

No. of atoms

= N_A $$ \times $$ No. of moles $$ \times $$ 3

= 6.023 $$ \times $$ 10²³ $$ \times $$ 0.1 $$ \times $$ 3

= 1.806 $$ \times $$ 10²³

H2	+	1/2O2	$$ \to $$	H2O
2 g		16 g		18 g
1 mol		0.5 mol		1 mol

10 g of H₂ = 5 mol and 64 g og O₂ = 2 mol

$$ \therefore $$ In this reaction, oxygen is the limiting reagent so amount of H₂O produced depends on that of O₂

Since 0.5 mol of O₂ gives 1 mol H₂O

$$ \therefore $$ 2 mol of O₂ will give 4 mol H₂O

C3H8	+	5O2	$$ \to $$	3CO2	+	4H2O	(balanced equation)
1 vol.		5 vol.		3 vol.		4 vol.

According to the above equation 1 vol. or 1 litre of propane requires to 5 vol. or 5 litre of O₂ to burn completely.