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If you add 5.3 g of sodium carbonate to excess dilute sulphuric acid, what mass of sodium sulphate would be made? ( na = 23 , c = 12 o = 16 16,532)
Accepted Answer
Here's how to calculate the mass of sodium sulfate produced:
1. Write the balanced chemical equation:
Na₂CO₃(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l) + CO₂(g)
2. Calculate the moles of sodium carbonate:
Moles of Na₂CO₃ = mass / molar mass = 5.3 g / (2
23 + 12 + 3
16) g/mol = 0.05 mol
3. Determine the mole ratio from the balanced equation:
The equation shows a 1:1 mole ratio between Na₂CO₃ and Na₂SO₄.
4. Calculate the moles of sodium sulfate produced:
Moles of Na₂SO₄ = 0.05 mol
5. Calculate the mass of sodium sulfate:
Mass of Na₂SO₄ = moles
molar mass = 0.05 mol
(2
23 + 32 + 4
16) g/mol = 7.1 g
Therefore, 7.1 g of sodium sulfate would be produced.
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