consider the reaction, 2Al(OH)3+3H2SO4---->Al2(SO4)3+6H2O a. Given that 152g of H2SO4 reacts with 234g of Al(OH)3 Howmany grams of Al2(SO4)3 will be theoretical generated? b. which of the reactants is the limiting reagent? c. If the actuall yield was 135 g Al(SO4)3, what is the percent yield?
I solve these limiting reagent problems by solving two simple stoichiometry problems, the first time with one reagent, the second time with the other reagent. The one providing the SMALLEST number of moles for the product will be the value to choose AND it is the limiting reagent. Here is a site that will show you in detail how to solve for each reagent.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Post your work if you get stuck.
To solve this problem, we need to determine the limiting reagent first, and then use stoichiometry to calculate the theoretical yield. Finally, we can calculate the percent yield using the actual yield and the theoretical yield.
a. To find the grams of Al2(SO4)3, we need to use the given quantities of H2SO4 and Al(OH)3 and consider the stoichiometric ratios from the balanced equation:
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Given:
Mass of H2SO4 = 152g
Mass of Al(OH)3 = 234g
First, we need to calculate the number of moles of each reactant:
Molar mass of H2SO4 = 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol
Number of moles of H2SO4 = mass (g) / molar mass (g/mol) = 152g / 98.09 g/mol = 1.55 mol
Molar mass of Al(OH)3 = 2(26.98) + 3(1.01) + 3(16.00) = 78.00 g/mol
Number of moles of Al(OH)3 = mass (g) / molar mass (g/mol) = 234g / 78.00 g/mol = 3.00 mol
Now, we compare the mole ratios from the balanced equation:
Mole ratio of H2SO4 to Al(OH)3 = 3 / 2 = 1.5
The mole ratio tells us that for every 1.5 moles of H2SO4, we need 1 mole of Al(OH)3. Since the ratio is less than 1, it means that Al(OH)3 is the limiting reagent.
To find the theoretical yield of Al2(SO4)3, we will use the mole ratio of Al(OH)3 to Al2(SO4)3, which is 1 / 1 (from the balanced equation). Therefore, the moles of Al(OH)3 and Al2(SO4)3 are equal.
Molar mass of Al2(SO4)3 = 2(26.98) + 3(32.07) + 12(16.00) = 342.15 g/mol
Theoretical yield (mass of Al2(SO4)3) = moles of Al2(SO4)3 x molar mass (g/mol)
= 3.00 mol x 342.15 g/mol = 1026.45 g
Therefore, the theoretical yield of Al2(SO4)3 is 1026.45 grams.
b. The limiting reagent is the reactant that is completely consumed (runs out) first. In this case, Al(OH)3 is the limiting reagent because its mole ratio is less than that of H2SO4.
c. The percent yield is calculated using the formula:
Percent yield = (actual yield / theoretical yield) x 100
Given:
Actual yield = 135g
Theoretical yield (calculated in part a) = 1026.45g
Percent yield = (135g / 1026.45g) x 100 ≈ 13.15%
Therefore, the percent yield is approximately 13.15%.
To solve this problem, we need to use the concept of stoichiometry, which relates the balanced chemical equation to the masses of the reactants and products.
a. To find the theoretical yield of Al2(SO4)3, we first need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed. Assuming all reactants are completely consumed, we can calculate the grams of Al2(SO4)3 produced using the given mass of H2SO4.
1. Calculate the number of moles of H2SO4:
- Using the molar mass of H2SO4 (98.09 g/mol), divide the given mass (152 g) by the molar mass to obtain moles:
moles of H2SO4 = 152 g / 98.09 g/mol
2. Use the balanced chemical equation to establish the stoichiometric ratio:
- From the balanced equation, we see that the ratio between H2SO4 and Al2(SO4)3 is 3:1.
3. Determine the moles of Al2(SO4)3:
- Multiply the moles of H2SO4 by the stoichiometric ratio:
moles of Al2(SO4)3 = moles of H2SO4 * (1 mole Al2(SO4)3 / 3 moles H2SO4)
4. Calculate the mass of Al2(SO4)3:
- Multiply the moles of Al2(SO4)3 by the molar mass of Al2(SO4)3 (342.15 g/mol):
mass of Al2(SO4)3 = moles of Al2(SO4)3 * 342.15 g/mol
b. To determine the limiting reagent, compare the moles of each reactant to the stoichiometric ratio. The reactant with fewer moles compared to the stoichiometric ratio is the limiting reagent.
c. To calculate the percent yield, divide the actual yield by the theoretical yield and multiply by 100:
percent yield = (actual yield / theoretical yield) * 100
Given that the actual yield is 135 g Al2(SO4)3, and to find the theoretical yield from part a, substitute these values into the equation and solve for the percent yield.
Note: It's important to have the molar masses of the compounds involved in the reaction to perform the necessary calculations.