How many grams of CaCo3 are required to prepare 50.0g of CaO

1 answer:

7 0

Answer is: 89 g of CaCO₃ is equired to prepare 50.0g of CaO.
Chemical reaction: CaCO₃ → CaO + CO₂.
m(CaO) = 50,0 g.
m(CaCO₃) = ?
n(CaO) = m(CaO) ÷ M(CaO).
n(CaO) = 50,0 g ÷ 56 g/mol.
n(CaO) = 0,89 mol.
from reaction: n(CaCO₃) : n(CaO) = 1 : 1.
n(CaCO₃) = n(CaO).
n(CaCO₃) = 0,89 mol.
m(CaCO₃) = m(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 0,89 mol · 100 g/mol
m(CaCO₃) = 89 g.
n - amount of substance.

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