Respuesta :
Answer:
Theoretical yield: 13.9 g
Percent yield: 94 %
Explanation:
The reaction is:
CaCO₃ → CaO + CO₂
We see that the reaction is correctly balanced.
1 mol of calcium carbonate can decompose to 1 mol of calcium oxide and 1 mol of carbon dioxide.
We convert the mass to moles: 24.8 g . 1mol / 100.08g = 0.248 moles
As ratio is 1:1, 0.248 moles of salt can decompose to 1 mol of oxide.
We convert the moles to mass: 0.248 mol . 56.08g /1mol = 13.9 g
That's the theoretical yield.
To determine the percent yield we think:
(Determined yield / Theoretical yield) . 100 → (13.1 / 13.9) . 100 = 94 %
The percent yield of Calcium oxide formed from calcium carbonate is 94 %.
How we calculate the percent yield?
Percent yield of any substance will be calculated as:
% yield = (Actual yield / Theoretical yield) × 100%
Given chemical reaction is:
CaCO₃ → CaO + CO₂
From the stoichiometry of the reaction, it is clear that moles of produced calcium oxide and reacted calcium carbonate is same i.e. they have 1:1 moles ratio.
Now we calculate the moles of CaCO₃ by using the below formula:
n = W/M, where
W = given mass = 24.8 grams
M = molar mass = 100.08g
Moles of CaCO₃ = 24.8 / 100.08 = 0.248 moles
So, produced moles of CaO is also = 0.248 moles
Now we calculate the mass of CaO by using the formula as:
W = n × M
Mass of CaO = 0.248mole × 56.08g /mole = 13.9 g
Given actual yield of CaO = 13.1 grams
Now we put required values on the % yield equation and we get:
% yield = (13.1/13.9) × 100 = 94%
Hence, 94% is the percent yield of CaO.
To know more about % yield, visit the below link:
https://brainly.com/question/8638404
