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How much caco3 is produced if 47.5 moles nh3 are produced?

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Answer is: 2380 grams of CaCO₃ is produced.
Chemical reaction: Ca(CN)₂ + 3H₂O → CaCO₃ + 2NH₃.
n(NH₃) = 47,5 mol.
From chemical reaction: n(CaCO₃) : n(NH₃) = 1 : 2.
n(CaCO₃) = 47,5 mol ÷ 2.
n(CaCO₃) = 23,75 mol.
m(CaCO₃) = n(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 23,75 mol · 100,09 g/mol.
m(CaCO₃) = 2380 g.
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