00:01
Hi, so to solve for this, let's write first the balance equation for this reaction.
00:06
We have barium chloride, that's bacl2, and potassium sulfate, that's k2so4, and then the products barium sulfate, that's baso4, and potassium chloride, kcl.
00:23
So we need to balance this first, we just need to write 2 in here.
00:27
This is the balance equation for this reaction.
00:29
Now we have to determine first the limiting reactant, we'll convert the given mass of reactants to moles.
00:35
We'll start with barium chloride, if we have 81 .6 grams of this reactant, convert this to moles by dividing the molar mass of barium chloride, that's 208 .23 grams, and then we have here moles, so we could cancel grams, and we'll get 0 .3919.
00:58
This is moles of barium chloride.
01:02
And then for k2so4, this is potassium sulfate, we have 71 .4 grams, divided by the molar mass of k2so4, that's 174 .26 grams per mole.
01:18
So we could cancel grams, and we'll get 0 .4097, this is moles of potassium sulfate.
01:28
And then we will solve for mole ratio.
01:33
Mole ratio is the number of moles of the reactant divided by the coefficient of that reactant from the balance equation.
01:39
So for barium chloride, we have 0 .3919 moles, and then coefficient of barium chloride is 1.
01:48
So divided by 1, this will still give us 0 .3919.
01:55
For k2so4, we have 0 .4097, and then the coefficient of k2so4 is also 1, so divided by 1, this will still give us 0 .4097.
02:11
And whichever gives the smallest mole ratio will be the limiting reactant.
02:15
So since 0 .3919 is less than 0 .4097, barium chloride is our limiting reagent here, our reactant.
02:25
So that means the amount of the product that will be formed from the reaction will be based on the number of moles of barium chloride, since it is the limiting reactant.
02:35
Now we could solve for the first question, what is the maximum amount of barium sulfate that can be formed? we'll calculate the mass of baso4 using the number of moles of barium chloride, that's 0 .3919...