Question

how much sulphuric acid is required to dissolve completely 3g of magnesium carbonate. calculate the weight of co2 liberated which converts sodium hydroxide into sodium carbonated. what is the weight of sodium carbonate form?

Answer 1

Consider the reaction between magnesium carbonate and sulphuric acid,
MgCO₃ + H₂SO₄ → MgSO₄ + H₂O + CO₂
Here, 1 mole of MgCO₃ reacts with 1 mole of H₂SO₄ to yield 1 mole of MgSO₄ and 1 mole CO₂
Given,
Mass of magnesium carbonate = 3 g
Hence,
$$\begin{gathered} Amountof{H}_{2}S{O}_{4}needed=\frac{3g}{84.3139g/mol}\times \frac{1mole{H}_{2}S{O}_4}{1moleMgC{O}_3} \\ =\mathbf{0}\mathbf{.}\mathbf{03558}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s} \end{gathered}$$

Now, consider the reaction of sodium hydroxide with the liberated 1 mole of CO₂ to form sodium carbonate,
Mass of CO₂ liberated = 1 mole x 44 g/mol = 44 g
The formation of sodium carbonate is as follows:
2NaOH + CO₂ → Na₂CO₃ + H₂O
Here, 2 moles of NaOH reacts with 1 mole of CO₂ to yield 1 mole of Na₂CO₃
Mass of NaOH needed = 2 moles x 40 g/mol = 80 g/mol
Therefore, as CO₂ is the limiting reagent, Na₂CO₃ will be formed from only that amount of available CO₂,
Hence,
Number of moles of Na₂CO₃ = 1 mole
Mass of Na₂CO₃ = 1 mole x 105.9888 g/mol = 105.9888 g