Question

What mass of caco3 is required to react completely with 25 ml of 0.75m hcl?

Answer 1

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

Answer 2

Answer:

Explanation:

Refer the below attachment.