When a mixture of sodium carbonate and magnesium carbonate was calcined to constant weight, 4.48 liters of gas were released. The solid residue was completely reacted with 73 g of a 25% hydrochloric acid solution. Calculate the mass fraction of sodium carbonate in the original mixture.
Let us write the equation for the thermal decomposition of magnesium carbonate:
MgCO3 → MgO + CO2 (1)
Thus, the solid residue is a mixture of the formed magnesium oxide and the initial sodium carbonate. Both components of the solid residue react with hydrochloric acid:
MgO + 2HCl → MgCl2 + H2O (2)
Na2CO3 + 2HCl → MgCl2 + CO2 ↑ + H2O (3)
Let us calculate the amount of substance released by CO2 released during the decomposition of MgCO3:
n (CO2) = 4.48 L / 22.4 L / mol = 0.2 mol.
From equation (1): n (MgO) = n (CO2) = 0.2 mol,
m (MgO) = n (MgO) ∙ M (MgO) = 0.2 mol ∙ 40 g / mol = 8 g.
Let’s find the amount of hydrochloric acid substance required for the reaction with MgO:
n (HCl) 2 = 2 ∙ n (MgO) = 2 ∙ 0.2 mol = 0.4 mol.
Let’s find the total mass and amount of the hydrochloric acid substance taken for the reaction:
m (HCl (total)) in-islands = m (HCl (total)) solution ∙ ω (HCl) = 73 g ∙ 0.25 = 18.25 g,
n (HCl (gen.)) = m (HCl (gen.)) in islands / M (HCl) = 18.25 g / 36.5 g / mol = 0.5 mol.
Let’s find the amount of hydrochloric acid substance required for the reaction with Na2CO3:
n (HCl) 3 = n (HCl (gen.)) – n (HCl) 2 = 0.5 mol – 0.4 mol = 0.1 mol.
Find the amount of substance and the mass of sodium carbonate in the original mixture.
From equation (3): n (Na2CO3) = 0.5 ∙ n (HCl) 3 = 0.5 ∙ 0.1 mol = 0.05 mol.
m (Na2CO3) = n (Na2CO3) ∙ M (Na2CO3) = 0.05 mol, ∙ 106 g / mol = 5.3 g.
Let’s find the amount of substance and the mass of magnesium carbonate in the initial mixture.
From equation (1): n (MgCO3) = n (CO2) = 0.2 mol,
m (MgCO3) = n (MgCO3) ∙ M (MgCO3) = 0.2 mol ∙ 84 g / mol = 16.8 g.
Determine the mass of the initial mixture and the mass fraction of sodium carbonate in it:
m (MgCO3 + Na2CO3) = m (MgCO3) + m (Na2CO3) = 16.8 g + 5.3 g = 22.1 g.
ω (Na2CO3) = m (Na2CO3) / m (MgCO3 + Na2CO3) = 5.3 g / 22.1 g = 0.24 (24%).
Answer: ω (Na2CO3) = 24%.
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