Find the limiting reactant for each initial amount of reactants.
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Question

Find the limiting reactant for each initial amount of reactants.
4 Al(s) + 3 O2( g)¡2 Al2O3(s)
a. 1 mol Al, 1 mol O2
b. 4 mol Al, 2.6 mol O2
c. 16 mol Al, 13 mol O2
d. 7.4 mol Al, 6.5 mol O2

Accepted Answer

Hi everyone today, we have a question telling us given the following chemical reaction, identify the limiting reactant based on each initial amount provided. So we're going to rewrite our chemical equation. We have aluminum solid plus oxygen gashes forms aluminum oxide solid. So if we notice on our react inside we have two oxygen, which means we need an even number. So we're going to fix that by putting a two in front of our aluminum oxide and we can put a three in front of our oxygen to balance that and then a four in front of our aluminum. And now we can start with our equations. So starting with a we have 15 moles of aluminum times are multiple ratio. So two moles of aluminum oxide over four moles of aluminum and our moles of aluminum are going to cancel out and that's going to equal 7.5 moles of aluminum oxide. And now for our oxygen we have 11 moles of oxygen. And we're going to multiply by our multiple ratio. So two moles of aluminum oxide over three moles of oxygen and our moles of oxygen are going to cancel out. And that's going to give us 7. moles of aluminum oxide and are less amount here is given to us by the oxygen. So our answer for a is oxygen. Now we're going to move on to be. So we have three moles of aluminum Times are multiple ratio. So two moles of aluminum oxide over four moles of aluminum and our moles of aluminum are going to cancel out And give us 1.5 moles of aluminum oxide. And then we have two moles of oxygen Times are multiple ratio. So two moles of aluminum oxide. Over three moles of oxygen equals 1. moles of aluminum oxide. And that gives us the lesser number. So our answer for B is oxygen. And now we're going to move on to see. So we have 3.7 moles of aluminum times are multiple ratio. So two moles of aluminum oxide Over four moles of aluminum And that equals 1. moles of aluminum oxide. And then we have 3.2 moles of oxygen. And we're gonna multiply that by our multiple ratio. So two moles of aluminum oxide over three moles of oxygen. And that gives us 2. moles of aluminum oxide. So our lesser amount here is the 1.85. So our answer for C is aluminum. And lastly we have D. So we have two moles of aluminum Times are multiple ratio. So two moles of aluminum oxide over four moles of aluminum equals one mole of aluminum oxide. And for oxygen, we have two moles of oxygen times two moles of aluminum oxide over three moles of oxygen. And that equals 1. moles of aluminum oxide. So that is our greater number. So our answer here is going to be aluminum as well. Thank you for watching. Bye

Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O2( g)¡2 Al2O3(s) a. 1 mol Al, 1 mol O2 b. 4 mol Al, 2.6 mol O2 c. 16 mol Al, 13 mol O2 d. 7.4 mol Al, 6.5 mol O2

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