00:01
This is a limiting reaction problem because i'm given the mass of each reactant, but i can't determine which is the limiting reactant without first converting to moles.
00:10
So i'll divide by the molar mass of each substance.
00:20
So this is mole.
00:22
F and ah.
00:27
I'll repeat for the second reactant.
00:41
Okay.
00:42
Notice that the reactant coefficients are 2 to 1.
00:48
So that means that i need twice as much sodium hydroxide as i do this for a compound, but i can already see that i don't have enough of the sodium hydroxide compound.
01:07
If you can't see that, what we will do is divide by two.
01:13
Or maybe instead of the biome 2, let's say multiply by one half for that multiple ratio.
01:20
And what we see is that if we use all of this hydroxide, we need 0 .178 moles of b2h6.
01:37
So if all of this is used according to the 1 to 2 ratio, i will need with 178.
01:47
But i have a way more than that.
01:50
I have an x set...