When excess dilute HCL was added to sodium sulphate, 960cm^3 of sulphuric(iv)oxide gas was produced. Calculate the mass of sodium sulphate that was used (Molar gas volume=24000cm^3 and molar mass of sulphate=126g)
Question
When excess dilute HCL was added to sodium sulphate, 960cm^3 of sulphuric(iv)oxide gas was produced. Calculate the mass of sodium sulphate that was used (Molar gas volume=24000cm^3 and molar mass of sulphate=126g)
Solution
The reaction between sodium sulphate and hydrochloric acid is as follows:
Na2SO4 + 2HCl -> 2NaCl + H2SO4
From the balanced chemical equation, we can see that one mole of sodium sulphate (Na2SO4) produces one mole of sulphuric acid (H2SO4).
Given that the molar volume of a gas at standard temperature and pressure (STP) is 24,000 cm^3, the 960 cm^3 of sulphuric acid gas produced corresponds to 960/24,000 = 0.04 moles of sulphuric acid.
Since the reaction is 1:1, this also means that 0.04 moles of sodium sulphate were used.
The molar mass of sodium sulphate (Na2SO4) is calculated as follows: (223) + (32) + 4(16) = 142 g/mol.
Therefore, the mass of sodium sulphate used is 0.04 moles * 142 g/mol = 5.68 g.
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