#6
If 10.0 grams of Aluminum carbonate(s) decomposes to produce aluminum oxide(s) and carbon dioxide(g), how many grams of CO2(g) are formed? If 5.00 grams is recovered experimentally what is the percent yield?
To find the number of grams of CO2 formed, we need to first determine the molar mass of Aluminum carbonate (Al2(CO3)3).
Aluminum (Al) has a molar mass of 26.98 g/mol, Carbon (C) has a molar mass of 12.01 g/mol, and Oxygen (O) has a molar mass of 16.00 g/mol.
So the molar mass of Aluminum carbonate is:
(2 * 26.98 g/mol) + (3 * (12.01 g/mol + 3 * 16.00 g/mol))
= 54.94 g/mol + 147.03 g/mol
= 201.97 g/mol
Next, we need to calculate the number of moles of Aluminum carbonate that decompose. We can do this by dividing the given mass (10.0 g) by the molar mass (201.97 g/mol):
10.0 g / 201.97 g/mol ≈ 0.049 moles
According to the balanced chemical equation,
Al2(CO3)3(s) → Al2O3(s) + 3CO2(g)
We can see that 1 mole of Aluminum carbonate produces 3 moles of CO2. So, 0.049 moles of Aluminum carbonate will produce:
0.049 moles × 3 = 0.147 moles of CO2
To find the mass of CO2 formed, multiply the number of moles by the molar mass of CO2 (which is 44.01 g/mol):
0.147 moles × 44.01 g/mol ≈ 6.47 g
Therefore, approximately 6.47 grams of CO2 gas will be formed.
Now, to calculate the percent yield of CO2, divide the actual yield (5.00 g) by the theoretical yield (6.47 g), and multiply by 100%:
(5.00 g / 6.47 g) × 100% ≈ 77.32%
So, the percent yield of CO2 is approximately 77.32%.