00:01
Hi, so to solve for this we need to write first the balance equation for this reaction.
00:05
We have n2 and h2, this will form ammonia and h3.
00:11
So we need to balance this first, we'll write 2 here and then we need to write 3 here.
00:15
And this is the balance equation for this reaction.
00:18
Since we're given with both amounts of the reactants, we have to convert the given mass of nitrogen gas and hydrogen gas to moles.
00:26
We'll start with n2, we have 5 .10, this is grams of n2, convert to moles by dividing the molar mass of n2, that's 28 .02 grams and then we have here moles.
00:38
So we could cancel grams and this will give us 0 .182, this is moles of nitrogen gas.
00:46
For hydrogen gas we have 4 .14 grams of h2, divide the molar mass, that's 2 .02 grams and then we have here moles, so that means we could cancel this one and this will give us 2 .05 moles of h2.
01:05
And we will solve for mole ratio.
01:09
Mole ratio is the number of moles of the reactant divided by the coefficient of that reactant from the balance equation.
01:15
So for n2 we have 0 .182 divided by coefficient of n2 which is 1, so this will give us 0 .182.
01:25
Also for h2 we have 2 .05 moles divided by the coefficient which is 3, so this will give us 0 .683.
01:36
And whichever gives the smallest mole ratio will be the limiting reactant.
01:40
So therefore nitrogen gas is the limiting reactant here and the amount of n2 will be the basis and the amount of your product that will be formed the reaction.
01:50
Now we could solve for the first question, the mass of ammonia using the number of moles of n2...