00:01
Okay, so in this problem, we're dealing with the reaction, 6 sodium plus iron oxide goes to form 3 sodium oxide plus 2 irons, and we're given 100 grams of starting material for sodium and iron.
00:16
So what we want to find out first for part a is our limiting reagent, and part b is our excess reagent.
00:23
We can do this simultaneously.
00:26
So what we're going to do first is transfer our 100 grams of sodium into.
00:31
Molecular units by multiplying by the mole ratio of one mole of sodium for every 22 .99 grams of sodium and we get 4 .35 moles of sodium.
00:42
Now we want to multiply by our reaction coefficient of six moles for every one reaction and our answer is 0 .725 reactions based on sodium.
00:53
Now we're going to calculate how many reactions we can perform with our 100 grams of iron oxide by multiplying again by 1 mole ratio, 1 mole for every 159 .69 .6 grams, and we get 0 .626 moles of iron oxide.
01:09
And we're going to multiply this again by our reaction ratio of one reaction for 1 mole.
01:14
And we're left with 0 .626 reactions based on iron oxide.
01:19
Since our sodium is greater, we can perform more reactions with our sodium...