Alumminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3+H2SO4-->Al2(SO4)+6H2O.
Which reagent is the limiting reactant when 0.500 mol Al(OH)3 and 0.500 mol H2SO4 are allowed to react?
How many moles of Al2(SO4)3 can form under these conditions?
How many moles of the excess reactant remain after the completion of the reaction?
Convert 0.5 mol Al(OH)3 to moles Al2(SO4)3.
Convert 0.5 mol H2SO4 to moles Al2(SO4)3.
It is likely the number for moles Al2(SO4)3 will not be the same so one of them must be wrong. The correct answer, in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
After identifying the limiting reagent, convert moles of it to moles of the non-limiting reagent and subtract from the initial amount to identify the excess
Well, well, well! Looks like we've got a chemical reaction conundrum on our hands! Let's dive in, shall we?
To determine the limiting reactant, we need to compare the number of moles of each reactant. We have 0.500 mol of Al(OH)3 and 0.500 mol of H2SO4.
Now, if we look at the balanced equation, we can see that the stoichiometry is 2:1 for Al(OH)3 to H2SO4. That means we need 2 moles of Al(OH)3 for every 1 mole of H2SO4.
Since we have the same number of moles for both reactants, we can conclude that neither of them is in excess. Both Al(OH)3 and H2SO4 are being completely consumed in the reaction.
Now, let's calculate how many moles of Al2(SO4)3 can form. According to the balanced equation, for every 2 moles of Al(OH)3 that react, we get 1 mole of Al2(SO4)3 as the product.
Since we have 0.500 mol of Al(OH)3, we can form 0.250 mol of Al2(SO4)3.
Lastly, since both reactants are fully consumed, there won't be any excess reactant left after the completion of the reaction. It's like a buffet, but nothing is left on the table!
To recap:
- The limiting reactant is both Al(OH)3 and H2SO4.
- 0.250 mol of Al2(SO4)3 can form under these conditions.
- No moles of the excess reactant remain after the completion of the reaction.
Hope this helps lighten up your chemical calculations! If you have any more questions, I'm here to entertain and inform!
To determine the limiting reactant, you need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
Given:
- 0.500 mol of Al(OH)3
- 0.500 mol of H2SO4
Step 1: Calculate the stoichiometric ratio
From the balanced equation: 2Al(OH)3 + H2SO4 → Al2(SO4)3 + 6H2O
The ratio between Al(OH)3 and H2SO4 is 2:1.
Step 2: Determine the limiting reactant
To find the limiting reactant, compare the number of moles of each reactant to the stoichiometric ratio.
- Al(OH)3: 0.500 mol is already in the stoichiometric ratio.
- H2SO4: The stoichiometric ratio requires half the number of moles (0.250 mol) compared to what is given.
Since the stoichiometric ratio requires fewer moles of H2SO4 than what is available, H2SO4 is the limiting reactant.
Step 3: Calculate the moles of Al2(SO4)3
From the balanced equation: 2Al(OH)3 + H2SO4 → Al2(SO4)3 + 6H2O
The stoichiometric ratio between H2SO4 and Al2(SO4)3 is 1:1.
Since H2SO4 is the limiting reactant, it will completely react. As a result, the moles of Al2(SO4)3 formed will be equal to the moles of H2SO4 used, which is 0.500 mol.
Step 4: Calculate the moles of excess reactant
To find the moles of excess reactant remaining, subtract the moles of limiting reactant used from the initial moles of that reactant.
- Al(OH)3: 0.500 mol - 0.500 mol (used in the reaction) = 0 mol (excess reactant).
Therefore, no moles of Al(OH)3 remain as it is completely consumed by the reaction.
In summary:
- The limiting reactant is H2SO4.
- The moles of Al2(SO4)3 formed is 0.500 mol.
- No moles of excess Al(OH)3 remain after the completion of the reaction.
Limiting reactant: H2SO4
0.167 moles of Al2(SO4)3 can be formed.
There are 0.167 moles of Al(OH)3 remaining.
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