00:01
Hi, in this question we are given that aluminum reacts with oxygen to produce aluminum oxide that is al2o3.
00:09
This is the balanced equation in which 4 moles of aluminum reacts with 3 moles of oxygen gas to produce 2 moles of aluminum oxide.
00:16
We are given with mass of aluminum and oxygen that is 2 .5 grams and we are asked to calculate the maximum amount of aluminum oxide that is produced in this reaction.
00:27
First of all we will calculate moles of aluminum and oxygen.
00:32
It will be given by mass over molar mass and we will take the atomic mass in case of aluminum.
00:39
Now we can substitute the values for both here.
00:43
In this manner now we can eliminate the common unit gram from here and on solving we are getting the moles of aluminum equals to 0 .093 mole and moles of oxygen gas equals to 0 .078 mole.
01:02
Now we need to identify the limiting reagent among the given ones.
01:06
We can say the limiting reagent is the one which is completely consumed in the reaction.
01:13
Here we can write aluminum and here we can write oxygen.
01:18
According to stoichiometry of reaction we can say that 4 moles of aluminum reacts with 3 moles of oxygen gas.
01:25
We are having 0 .093 moles of aluminum.
01:29
It will react with 3 mole divided by 4 mole into 0 .093 moles of oxygen gas.
01:40
On solving we are getting 0 .070 moles of oxygen gas but we are having 0 .078 moles of oxygen gas...