If 852.04 g Al(NO3)3 reacted with the 741.93 g Na2CO3, which would be the limiting reagent?
c. Use the limiting reagent to determine how many grams of Al2(CO3)3 should precipitate out in the reaction.
d. If only 341.63 g of Al2(CO3)3 precipitate were actually collected from the reaction, what would the percent yield be?
To determine the limiting reagent, we need to compare the moles of Al(NO3)3 and Na2CO3 and see which one is present in a lesser amount relative to the stoichiometry of the balanced equation.
The balanced equation for the reaction between Al(NO3)3 and Na2CO3 is:
2 Al(NO3)3 + 3 Na2CO3 → Al2(CO3)3 + 6 NaNO3
1. Calculate the moles of Al(NO3)3:
Molar mass of Al(NO3)3 = atomic mass of Al + (atomic mass of N * 3) + (atomic mass of O * 9)
= 26.98 g/mol + (14.01 g/mol * 3) + (16.00 g/mol * 9)
≈ 212.00 g/mol
Moles of Al(NO3)3 = mass / molar mass
= 852.04 g / 212.00 g/mol
≈ 4.02 mol
2. Calculate the moles of Na2CO3:
Molar mass of Na2CO3 = (atomic mass of Na * 2) + atomic mass of C + (atomic mass of O * 3)
= (22.99 g/mol * 2) + 12.01 g/mol + (16.00 g/mol * 3)
≈ 105.99 g/mol
Moles of Na2CO3 = mass / molar mass
= 741.93 g / 105.99 g/mol
≈ 7.00 mol
3. Now, we will compare the moles of Al(NO3)3 and Na2CO3 to determine the limiting reagent. The ratio in the balanced equation is 2:3 (2 moles of Al(NO3)3 react with 3 moles of Na2CO3).
Since the ratio of Al(NO3)3 to Na2CO3 is 2:3, for every 2 moles of Al(NO3)3, we need 3 moles of Na2CO3.
If we have 4.02 moles of Al(NO3)3, we would need (4.02 mol * 3 mol/2 mol) = 6.03 moles of Na2CO3.
However, we only have 7.00 moles of Na2CO3, which is greater than the required amount of 6.03 moles. This means that Na2CO3 is in excess, and Al(NO3)3 is the limiting reagent.
c. To determine how many grams of Al2(CO3)3 precipitate out in the reaction, we need to use the stoichiometry of the balanced equation.
From the balanced equation, we know that the mole ratio of Al(NO3)3 to Al2(CO3)3 is 2:1.
Moles of Al2(CO3)3 = moles of Al(NO3)3 (from the limiting reagent)
Moles of Al2(CO3)3 = 4.02 mol
Molar mass of Al2(CO3)3 = (atomic mass of Al * 2) + (atomic mass of C) + (atomic mass of O * 3)
= (26.98 g/mol * 2) + 12.01 g/mol + (16.00 g/mol * 3)
≈ 233.98 g/mol
Mass of Al2(CO3)3 = moles of Al2(CO3)3 * molar mass of Al2(CO3)3
= 4.02 mol * 233.98 g/mol
≈ 940.96 g
Therefore, approximately 940.96 grams of Al2(CO3)3 should precipitate out in the reaction.
d. To calculate the percent yield, we need to use the actual yield and the theoretical yield.
Theoretical yield = mass of Al2(CO3)3 calculated in part c = 940.96 g
Actual yield = 341.63 g (given)
Percent yield = (actual yield / theoretical yield) * 100
= (341.63 g / 940.96 g) * 100
≈ 36.30%
Therefore, the percent yield of Al2(CO3)3 is approximately 36.30%.
To determine the limiting reagent, you need to compare the amount of each reactant with the stoichiometric ratio in the balanced chemical equation.
a. Convert the given masses of Al(NO3)3 and Na2CO3 to moles.
- The molar mass of Al(NO3)3 is 213.03 g/mol.
- The molar mass of Na2CO3 is 105.99 g/mol.
Moles of Al(NO3)3 = 852.04 g / 213.03 g/mol
Moles of Na2CO3 = 741.93 g / 105.99 g/mol
b. Determine the mole ratio between Al(NO3)3 and Na2CO3 from the balanced chemical equation.
The balanced chemical equation for the reaction is:
2 Al(NO3)3 + 3 Na2CO3 -> Al2(CO3)3 + 6 NaNO3
From the balanced equation, the mole ratio between Al(NO3)3 and Na2CO3 is 2:3.
c. Calculate the moles of Al2(CO3)3 that should be formed based on the limiting reagent.
To determine the limiting reagent, compare the moles of each reactant using the ratio from the balanced equation. The reactant that produces the smallest amount of product is the limiting reagent.
Moles of Al2(CO3)3 from Al(NO3)3 = (Moles of Al(NO3)3) * (1 mol Al2(CO3)3 / 2 mol Al(NO3)3)
Moles of Al2(CO3)3 from Na2CO3 = (Moles of Na2CO3) * (1 mol Al2(CO3)3 / 3 mol Na2CO3)
Whichever value is smaller is the actual moles of Al2(CO3)3 formed.
d. Calculate the mass of Al2(CO3)3 formed based on the limiting reagent.
Mass of Al2(CO3)3 = (Moles of Al2(CO3)3) * (Molar mass of Al2(CO3)3)
e. Calculate the percent yield.
Percent yield = (Actual mass of Al2(CO3)3 collected / Theoretical mass of Al2(CO3)3) * 100
Have I given you this page before. It will work all of your stoichiometry problems. For limiting reagent problems, just work the stoichiometry problem for each reactant, then the one producing the smaller number of mols of the product is the limiting reagent.
http://www.jiskha.com/science/chemistry/stoichiometry.html