5.00g aluminum sulfide and 10.0g water react according to the reaction:
Al2S3 + 6 H2O -> 2 Al(OH)3 + 3 H2S
What is maximum mass of H2S that can be formed?
What mass of excess reactant remains when the reaction is complete?
determinine the limiting reagent.
determine the moles of aluminum sulfide
determine the moles of water?
are there greater then 6x moles water than aluminum sulfide? If so, aluminum sulfide is the limiting reageant.
If aluminumsulfide is the limiting reagent, then you will get three times those moles of H2S. Convert to grams.
To find the maximum mass of H2S that can be formed, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction, thereby determining the maximum amount of product that can be formed.
1. Calculate the moles of aluminum sulfide (Al2S3) and water (H2O) using their respective molar masses:
Moles of Al2S3 = 5.00g / (2 * 26.98g/mol + 3 * 32.07g/mol) = 0.0183 mol
Moles of H2O = 10.0g / (2 * 1.007g/mol + 16.00g/mol) = 0.556 mol
2. Use the stoichiometric coefficients from the balanced chemical equation to find the mole ratio between Al2S3 and H2S:
From the balanced equation, 1 mole of Al2S3 produces 3 moles of H2S.
3. Determine the limiting reactant:
Since we have 0.0183 moles of Al2S3 and 0.556 moles of H2O, we can compare their mole ratios to find the limiting reactant.
The mole ratio of Al2S3 to H2S is 1:3, so the maximum moles of H2S that can be formed from the 0.0183 moles of Al2S3 is:
0.0183 mol Al2S3 * 3 mol H2S / 1 mol Al2S3 = 0.055 moles
Since 0.055 is less than 0.556, the moles of H2S we can obtain from water, Al2S3 is the limiting reactant.
4. Calculate the maximum mass of H2S that can be formed from Al2S3:
Mass of H2S = Moles of H2S * Molar mass of H2S
= 0.055 mol * (2 * 1.007g/mol + 32.07g/mol)
= 2.96g
So, the maximum mass of H2S that can be formed is 2.96g.
To determine the mass of the excess reactant remaining when the reaction is complete, we need to calculate the mass of water consumed during the reaction.
5. Calculate the moles of water consumed using the mole ratio between Al2S3 and H2O:
Moles of H2O consumed = Moles of Al2S3 * (6 moles H2O / 1 mole Al2S3)
= 0.0183 mol * 6
= 0.110 mol
6. Calculate the mass of water consumed:
Mass of H2O consumed = Moles of H2O consumed * Molar mass of H2O
= 0.110 mol * (2 * 1.007g/mol + 16.00g/mol)
= 3.55g
7. Finally, calculate the mass of excess reactant remaining:
Mass of excess H2O = Initial mass of H2O - Mass of H2O consumed
= 10.0g - 3.55g
= 6.45g
Therefore, when the reaction is complete, there will be 6.45g of excess water remaining.
To determine the maximum mass of H2S that can be formed, we need to calculate the amount of moles of aluminum sulfide and water and compare them.
1. Calculate the number of moles of aluminum sulfide (Al2S3):
- Molar mass of Al2S3 = (2 x 26.98 g/mol) + (3 x 32.06 g/mol) = 150.16 g/mol
- Moles of Al2S3 = Mass of Al2S3 / Molar mass of Al2S3
= 5.00 g / 150.16 g/mol
2. Calculate the number of moles of water (H2O):
- Molar mass of H2O = (2 x 1.01 g/mol) + (16.00 g/mol) = 18.02 g/mol
- Moles of H2O = Mass of H2O / Molar mass of H2O
= 10.0 g / 18.02 g/mol
3. Determine the limiting reactant:
The reactant that produces the least amount of product is the limiting reactant. In this case, to react with 6 moles of water, 1 mole of Al2S3 is needed. Therefore, the limiting reactant is aluminum sulfide.
4. Calculate the maximum moles of H2S that can be formed from Al2S3:
From the balanced equation, we see that 1 mole of Al2S3 produces 3 moles of H2S.
- Moles of H2S = 3 x Moles of Al2S3
5. Calculate the maximum mass of H2S:
- Mass of H2S = Moles of H2S x Molar mass of H2S
To determine the mass of excess reactant remaining, we need to calculate the moles of the limiting reactant used and subtract it from the initial moles of the excess reactant.
6. Calculate the moles of the excess reactant used (water):
From the balanced equation, we see that 1 mole of Al2S3 reacts with 6 moles of H2O.
- Moles of H2O used = Moles of Al2S3 x 6
7. Calculate the moles of the excess reactant remaining:
- Moles of H2O remaining = Initial moles of H2O - Moles of H2O used
8. Calculate the mass of the excess reactant remaining (water):
- Mass of H2O remaining = Moles of H2O remaining x Molar mass of H2O
Now, you can substitute the values and perform the calculations to find both the maximum mass of H2S formed and the mass of the excess reactant remaining.
This is a limiting reagent (LR) and you know that because values are given for BOTH reactants.
Al2S3 + 6 H2O -> 2 Al(OH)3 + 3 H2S
mols Al2S3 = grams/molar mass = ?
mols H2O = grams/molar mass = ?
Convert mols Al2S3 to mols H2S
Convert mols H2O to mols H2S
It is likely these two values will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Now take the smaller value and convert to grams H2S. g = mols x molar mass.