Calcium cyanamide, CaCN2, reacts with water to form calcium carbonate and ammonia. CaCN2(s)+3H2O(l)→CaCO3(s)+2NH3(g). How many g

Question

3 years ago

6

rams of CaCO3 form if 161 g of H2O reacts

1 answer:

Norma-Jean [14] · Answer 1 · 2022-05-09T10:25:30+03:00

298 g of calcium carbonate CaCO₃

Explanation:

We have the following chemical reaction:

CaCN₂ (s) + 3 H₂O (l) → CaCO₃ (s)+ 2 NH₃ (g)

number of moles = mass / molar weight

number of moles of H₂O = 161 / 18 = 8.94 moles

Knowing the chemical reaction we devise the following reasoning:

if 3 moles of H₂O produces 1 mole of CaCO₃

then 8.94 moles of H₂O produces X moles of CaCO₃

X = (8.94 × 1) / 3 = 2.98 moles of CaCO₃

mass = number of moles × molar weight

mass of CaCO₃ = 2.98 × 100 = 298 g

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