In a reaction vessel, 2.4 mol of Al(OH)3 and 5.3 mol of H2SO4 react.
Products: Al2(SO4)3, H2O
[I know...]
Moles of Al2(SO4)3 in container:1.2mol
Moles of H2O in container:7.2 mol
[But not...]
Moles of excess reactant in container:____mol
how do i find it?
You are correct that Al(OH)3 is the limiting reagent. I don't know how you calculated that but that is the right answer. After you know the limiting reagent (LR), the remainder of the problem starts there. You solve it as you would any stoichiometry problem.
2.4 moles Al(OH)3 x (3 moles H2SO4/2 moles Al(OH)3) = 2.4 x (3/2) = 3.6 moles H2SO4 used; therefore, 5.3-3.6 = moles remaining.
You could have used the Al2(SO4)3 to do it also.
1.2 moles Al2(SO4)3 x [3 moles H2SO4/1 mole Al2(SO4)3] = 1.2*3 = 3.6 moles H2SO4 used.
To find the moles of the excess reactant in the reaction, you need to determine the limiting reactant first. The limiting reactant is the reactant that will be completely consumed, while the excess reactant will be leftover after the reaction is complete.
To find the limiting reactant, you need to compare the mole ratios of the reactants. The balanced chemical equation for the reaction is:
2Al(OH)3 + 3H2SO4 -> Al2(SO4)3 + 6H2O
From the balanced equation, you can see that the mole ratio between Al(OH)3 and H2SO4 is 2:3. This means that for every 2 moles of Al(OH)3, you need 3 moles of H2SO4.
Given that you have 2.4 mol of Al(OH)3 and 5.3 mol of H2SO4, you can set up a ratio to compare the amounts:
(2.4 mol Al(OH)3) / 2 = (x mol H2SO4) / 3
Cross-multiplying, you get:
2.4 * 3 = 2 * x
7.2 = 2x
x = 7.2 / 2
x = 3.6 mol
Based on the calculation, you find that you would need 3.6 mol of H2SO4 to react completely with 2.4 mol of Al(OH)3. However, since you have 5.3 mol of H2SO4, it is the excess reactant.
Therefore, the moles of excess reactant in the container is 5.3 mol.
To find the moles of the excess reactant in the container, you need to first determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
In this case, you can find the limiting reactant by calculating the moles of Al(OH)3 that would react with the given amount of H2SO4, and the moles of H2SO4 that would react with the given amount of Al(OH)3.
1. Start by writing and balancing the chemical equation for the reaction:
Al(OH)3 + 3H2SO4 -> Al2(SO4)3 + 6H2O
2. Calculate the number of moles of H2SO4 that would react with 2.4 mol of Al(OH)3:
2.4 mol Al(OH)3 x (3 mol H2SO4 / 1 mol Al(OH)3) = 7.2 mol H2SO4
3. Compare this calculated value (7.2 mol) to the actual amount of H2SO4 given (5.3 mol). Since 5.3 mol is less than 7.2 mol, H2SO4 is the limiting reactant.
4. Once you determine the limiting reactant, you can find the moles of excess reactant by subtracting the moles of the limiting reactant from the initial moles of the excess reactant.
In this case, since H2SO4 is the limiting reactant, you start with 5.3 mol of H2SO4 and subtract the moles of reacting H2SO4 (which is 5.3 mol based on the balanced equation):
Moles of excess reactant = 5.3 mol H2SO4 - 5.3 mol H2SO4 = 0 mol
Therefore, the moles of excess reactant in the container is 0 mol.