Al_S=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
Here is a worked example of a limiting reagent problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
To determine the limiting reagent, we need to compare the number of moles of each reactant with their stoichiometric ratio in the balanced equation.
First, let's find the number of moles of Al:
Given mass of Al = 10 grams
Molar mass of Al = 26.98 g/mol
Number of moles of Al = (mass of Al) / (molar mass of Al)
Number of moles of Al = 10 g / 26.98 g/mol ≈ 0.371 moles
Next, let's find the number of moles of S:
Given mass of S = 15 grams
Molar mass of S = 32.07 g/mol
Number of moles of S = (mass of S) / (molar mass of S)
Number of moles of S = 15 g / 32.07 g/mol ≈ 0.468 moles
Now, let's look at the balanced equation:
Al + S -> Al2S3
From the balanced equation, the stoichiometric ratio between Al and S is 1:1. This means that 1 mole of Al reacts with 1 mole of S to form 1 mole of Al2S3.
Comparing the moles of Al and S, we can see that Al has fewer moles (0.371 moles) compared to S (0.468 moles). Therefore, Al is the limiting reagent.
To find the amount of Al2S3 formed, we need to determine the moles of Al2S3 produced based on the limiting reagent (Al). Since Al is in a 1:1 stoichiometric ratio with Al2S3, the number of moles of Al2S3 formed will be the same as the number of moles of Al.
Therefore, the amount of Al2S3 formed in the reaction is approximately 0.371 moles.
The excess reagent is the reactant that is not fully consumed in the reaction. In this case, the excess reagent is S, as it has more moles (0.468 moles) compared to the stoichiometric ratio with Al.
To find the amount of excess reagent left over after the reaction, we subtract the moles of the limiting reagent consumed from the initial moles of the excess reagent.
Initial moles of S = 0.468 moles
Moles of S consumed = moles of Al consumed (since they are in a 1:1 stoichiometric ratio)
Moles of S consumed = 0.371 moles
Excess moles of S remaining = Initial moles of S - Moles of S consumed
Excess moles of S remaining = 0.468 moles - 0.371 moles ≈ 0.097 moles
Therefore, approximately 0.097 moles of S are leftover after the reaction.