#6.
If 10.0g of Aluminum carbonate(s) decomposes to produce aluminum oxide(s) and carbon dioxide(g), how many grams of CO2(g) are formed? If 5.00g is recovered experimentally what is the % yield?
Look at your first post; this problem is almost identical.
To determine the grams of CO2 produced from the decomposition of aluminum carbonate, we first need to calculate 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 (Al2(CO3)3) can be calculated as follows:
2(Al) + 3(C) + 9(O) = 2(26.98 g/mol) + 3(12.01 g/mol) + 9(16.00 g/mol) = 233.99 g/mol
Next, we need to find the moles of aluminum carbonate. We are given that we have 10.0 g of aluminum carbonate.
Moles = mass / molar mass = 10.0 g / 233.99 g/mol = 0.0427 mol
According to the balanced chemical equation, 1 mole of aluminum carbonate decomposes to produce 3 moles of CO2. So, the moles of CO2 produced can be calculated as follows:
Moles of CO2 = 3 * Moles of aluminum carbonate = 3 * 0.0427 mol = 0.1281 mol
Now we can calculate the grams of CO2 formed by multiplying the moles of CO2 by its molar mass:
Mass of CO2 = Moles of CO2 * Molar mass of CO2 = 0.1281 mol * 44.01 g/mol = 5.64 g
Therefore, theoretically, 5.64 g of CO2 should be formed.
To calculate the percent yield, we use the formula:
Percent yield = (Actual yield / Theoretical yield) * 100
In this case, the actual yield is given as 5.00 g. So, substituting the values:
Percent yield = (5.00 g / 5.64 g) * 100 = 88.76%
The percent yield is approximately 88.76%.