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Grade 11Algebra

Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below.
CaCO (s) 2HCl(aq)------> CaCl (aq) CO2 + H2O
What volume of CO2 will be formed at STP when 54.75 g HCl reacts with 500 g of CaCO3? Name the limiting reagent. Calculate the number of moles left of excess reactant.

Profile image of Naveen Kumar
6 Years agoGrade 11
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1 Answer

Profile image of Arun
6 Years ago
1000 mL of 0.75 M HCl have 0.75 mol of HCl = 0.75×36.5 g = 24.375 g
∴ Mass of HCl in 25mL of 0.75 M HCl = 24.375/1000 × 25 g = 0.6844 g
From the given chemical equation,
CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
2 mol of HCl i.e. 73 g HCl react completely with 1 mol of CaCO3 i.e. 100g
∴ 0.6844 g HCl reacts completely with CaCO3 = 100/73 × 0.6844 g = 0.938 gm

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