Al+S=Al2S3
Balanced equ= 2Al+3S=Al2S3
10 grams of Al were mixed with 15 gram of S
Find: limiting reagent al or s?
How much Al2S3 were formed in reaction?
What is the excess reagent?Al or S?
How much excuss reagent left over after reactin?
Thanks! <3
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To determine the limiting reagent, we need to compare the moles of Al and S.
1. Calculate the moles of Al and S:
Molar mass of Al = 26.98 g/mol
Molar mass of S = 32.07 g/mol
Moles of Al = mass / molar mass = 10 g / 26.98 g/mol = 0.370 mol
Moles of S = mass / molar mass = 15 g / 32.07 g/mol = 0.468 mol
2. Determine the stoichiometric ratio:
From the balanced equation 2Al + 3S = Al2S3, we see that the ratio of Al to S is 2:3.
3. Compare the moles to find the limiting reagent:
Since the stoichiometric ratio is 2:3, we need twice as many moles of Al as S. However, we have fewer moles of Al (0.370 mol) compared to S (0.468 mol). Therefore, Al is the limiting reagent.
4. Calculate the amount of Al2S3 formed:
The balanced equation provides a 1:1 molar ratio for Al2S3 and Al. So, 0.370 mol of Al will react completely to form 0.370 mol of Al2S3.
Molar mass of Al2S3 = (2 x molar mass of Al) + (3 x molar mass of S)
= (2 x 26.98 g/mol) + (3 x 32.07 g/mol)
= 51.96 g/mol + 96.21 g/mol
= 148.17 g/mol
Mass of Al2S3 formed = moles x molar mass
= 0.370 mol x 148.17 g/mol
= 54.81 g
Therefore, 54.81 grams of Al2S3 were formed in the reaction.
5. Determine the excess reagent:
Since Al is the limiting reagent, S is the excess reagent.
6. Calculate the amount of excess reagent remaining:
To determine the amount of excess reagent left over, we need to calculate the moles of the excess reagent (S), subtract the moles used in the reaction, and then convert it back to grams.
Moles of S used in reaction = (moles of Al) x (3 moles of S / 2 moles of Al)
= 0.370 mol x (3/2)
= 0.555 mol
Moles of S remaining = Moles of S initial - Moles of S used in reaction
= 0.468 mol - 0.555 mol
= -0.086 mol (negative value means excess reagent is fully consumed)
Mass of excess S remaining = moles x molar mass
= (-0.086 mol) x (32.07 g/mol)
= -2.76 g (negative value means fully consumed)
Therefore, there is no excess S remaining after the reaction.
To summarize:
- The limiting reagent is Al.
- 54.81 grams of Al2S3 were formed in the reaction.
- The excess reagent is S.
- There is no excess S remaining after the reaction.
To determine the limiting reagent, we need to compare the number of moles of Al and S and see which reactant is used up first in the reaction.
1. Calculate the number of moles of Al and S:
- Moles of Al = Mass of Al / molar mass of Al
- Moles of S = Mass of S / molar mass of S
Molar mass of Al = 26.98 g/mol
Molar mass of S = 32.07 g/mol
Moles of Al = 10 g / 26.98 g/mol = 0.37 mol
Moles of S = 15 g / 32.07 g/mol = 0.47 mol
2. Examine the stoichiometry of the balanced equation:
From the balanced equation 2Al + 3S = Al2S3, we can see that the stoichiometric ratio is 2:3 for Al:S.
3. Determine the limiting reagent:
The stoichiometric ratio tells us that 2 moles of Al react with 3 moles of S to form 1 mole of Al2S3. Therefore, if the ratio of moles of Al to moles of S is less than the stoichiometric ratio, Al is the limiting reagent. In this case, we have a ratio of 0.37 mol Al to 0.47 mol S, which is indeed less than 2:3. Therefore, Al is the limiting reagent.
4. Calculate the amount of Al2S3 formed:
Since Al is the limiting reagent, we know that the moles of Al2S3 formed will be equal to the moles of Al. From step 1, we determined that the moles of Al is 0.37 mol. Therefore, 0.37 moles of Al2S3 were formed in the reaction.
5. Determine the excess reagent and the amount left over:
To find the excess reagent, we need to compare the moles of the non-limiting reagent (S) to the stoichiometric ratio. In this case, we have 0.47 mol of S and the stoichiometric ratio is 2:3. Therefore, we can determine that there is excess S.
Now, to find the amount of excess reagent left over, we need to calculate how much S reacted with the limiting reagent Al. Since the stoichiometry of the balanced equation is 2Al + 3S = Al2S3, the ratio of moles of Al to moles of S consumed is 2:3. From the limiting reagent calculation, we know that 0.37 mol of Al reacted. Therefore, the corresponding amount of S consumed would be (0.37 mol Al) / (2 mol Al : 3 mol S) = 0.2467 mol S.
To find the excess S left over, subtract the amount consumed by the reaction from the initial amount of S:
Excess S left over = Initial moles of S - Moles of S consumed
= 0.47 mol - 0.2467 mol
= 0.2233 mol
So, the excess reagent is S, and there is 0.2233 mol of excess S remaining after the reaction.