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lizisapenguin
24.11.2020 •
Chemistry
When heated, calcium carbonate decomposes according to the equation given below:
CaCO3 --> CaO + CO2
If 50.0 g of CaCO3 react but only 20.0 g of CO2 are recovered, what is the percent yield of this reaction?
90.9%
40.0%
66.7%
250%
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Ответ:
Percent yield = 90.9%
Explanation:
Given data:
Mass of CaCO₃ = 50.0 g
Mass of CO₂ produced = 20.0 g
Percent yield = ?
Solution:
Chemical equation:
CaCO₃ → CaO + CO₂
Number of moles of CaCO₃:
Number of moles = mass/molar mass
Number of moles = 50.0 g/ 100.1 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of CO₂ with CaCO₃.
CaCO₃ : CO₂
1 : 1
0.5 : 0.5
Mass of CO₂: Theoretical yield
Mass = number of moles × molar mass
Mass = 0.5 mol × 44 g/mol
Mass = 22 g
Percent yield:
Percent yield = ( actual yield / theoretical yield ) × 100
Percent yield = (20.0 g/ 22.0 g) × 100
Percent yield = 0.909 × 100
Percent yield = 90.9%
Ответ:
Hi. I'll explain
Explanation:
If you round it it will be 251
D is the answer (251)