Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
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1 Answers
10.75 M of HCl ≡ 0.75 mol of HCl are present in 1 L of water
≡ [(0.75 mol) × (36.5 g mol–1)] HCl is present in 1 L of water
≡ 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains 27.375 g of HCl.
Amount of HCl present in 25 mL of solution, by calculation
= 0.6844 g
From the given equation,
2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO3 (100 g).
Amount of CaCO3 that will react with 0.6844 g
= 0.9639 g