A mass of #2.40*g# of calcium oxide is obtained by heating a mass of #5.00*g# calcium carbonate. What is the stoichiometric equation that represents the formation of calcium oxide, and what is the percentage yield?
We need a stoichiometrically balanced equation to represent the decomposition of calcium carbonate..........
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Keep in mind that most carbonates react to heat in this manner:
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The stoichiometric equation for the formation of calcium oxide from calcium carbonate is:
CaCO₃(s) → CaO(s) + CO₂(g)
The molar mass of CaCO₃ is 100.09 g/mol, and the molar mass of CaO is 56.08 g/mol.
The theoretical yield of calcium oxide can be calculated using stoichiometry:
1 mol of CaCO₃ produces 1 mol of CaO
Therefore, the theoretical yield of CaO is:
(5.00 g CaCO₃) / (100.09 g/mol CaCO₃) * (56.08 g/mol CaO) = 2.80 g CaO (theoretical yield)
The actual yield is given as 2.40 g.
Percentage yield = (actual yield / theoretical yield) * 100
Percentage yield = (2.40 g / 2.80 g) * 100 = 85.71%
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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