SELECT * FROM question_mgmt as q WHERE id=1250 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=42 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

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?

Answer:

0.75 M of HCl ≡ 0.75 mol of HCl X molecular weight of HCl dissolved in 1000 ml of water

Or

[(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 Hcl = 27.375g

Or

1 ml of solutions contains Hcl = 27.375/1000 * 1

And 25 ml of solutions contains Hcl = 27.375/1000 * 25 = 0.6844 g.

From the given chemical equation,

CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

 

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

 

 

 

 

 


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