When excess dilute hydrochloric acid was added to sodium carbonate,960cm³ of carbon(IV)oxide gas was to produced.Calculate the mass of sodium carbonate that was used(Molar mass of sodium carbonate=106g and molar gas volume=24000cm³)
The balanced chemical equation for the reaction between hydrochloric acid and sodium carbonate is:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
From the equation, we can see that 2 moles of HCl react with 1 mole of Na2CO3 to produce 1 mole of CO2.
Given that 960 cm³ of CO2 gas was produced, we first need to convert this volume to moles using the molar gas volume:
1 mole of gas occupies 24000 cm³, therefore:
960 cm³ of CO2 is equal to (960/24000) = 0.04 moles of CO2
Since the ratio of HCl to Na2CO3 in the balanced equation is 2:1, we can conclude that 0.04 moles of CO2 were produced from (0.04 x 1/2) = 0.02 moles of Na2CO3.
The molar mass of Na2CO3 is given as 106 g/mol, therefore:
Mass of Na2CO3 used = molar mass x moles
Mass of Na2CO3 used = 106 g/mol x 0.02 mol
Mass of Na2CO3 used = 2.12 g
Therefore, the mass of sodium carbonate used is 2.12 grams.