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

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Flakey

Chemistry

6 October, 08:32

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

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Answers (1)

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Aaliyah Vance · Answer 1

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.