A49.3 sample of caco3 was treated with aqueous h2so4, producing calcium sulfate, 3.65 g of water and co2(g). what was the % yield of h2o?

41.1%

Explanation:

First write the balanced reaction:

CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Now calculate the theoretical yield:

49.3 g CaCO₃ × (1 mol CaCO₃ / 100 g CaCO₃) = 0.493 mol CaCO₃

0.493 mol CaCO₃ × (1 mol H₂O / 1 mol CaCO₃) = 0.493 mol H₂O

0.493 mol H₂O × (18 g H₂O / mol H₂O) = 8.87 g H₂O

Now calculate the % yield:

3.65 g H₂O / 8.87 g H₂O × 100% = 41.1%

Explanation:

MM: 100.09                                   18.02

       CaCO₃ + H₂SO₄ ⟶ CaSO₄ + H₂O + CO₂

m/g:  49.3                                        3.65  

1. Theoretical yield

(a) Moles of CaCO₃

(b) Moles of H₂O

(c) Theoretical mass of H₂O

(d) Percent yield

thats how its supposed to be