Chemistry Question?

Question

I am needing someones help badly. I am getting ready to take a chemistry test and I can't do this section at all. It is on theortical yield. Can somene please help me out! Here is a couple of questions.

How many moles of SiH4 could be formed from 19.09 moles of Mg2Si and 6.07 moles of water in the following reaction?
Mg2Si + 4 H20 ----------- 2 Mg(OH)2 + SiH4.

How many moles of Ba3(PO4) 2 could be formed from 6.76 moles of Ba(NO3)2 and 9.55 moles of Na3PO4 in the following reaction?
3 Ba(NO3) 2 + 2 Na3PO4 ------------------Ba3(PO4)2 + 6 NaNO3.

Any help would greatly be appreciated. Thanks.

Answer 1

Alrighty, here goes.

I'm going to explain the thought process behind solving these problems so you can hopefully use it to solve the questions on the test.

Basically, what you want to do is to find your limiting reactant. In your equation, you need to find out what on the left will run out first (once you run out, no more reaction...so you can figure out total quantity of products = theoretical yield). So, what we're going to do is:
1. Organize what you know. In the first example, you know that you have 19.09 moles of Mg2Si and 6.07 moles of H2O and that they react together by the given equation.

2. Figure out what you want to know. In a theoretical yield problem, you want to know the maximum possible product that could be formed. In this case, we want to know the maximum amount of SiH4 that can be formed.

3. Determine your strategy. Well, the amount of SiH4 that we can make is limited by how much of the starting materials we have. So we need to figure out which one runs out first. We know how much of each reactant we have, so we can use this information in several different ways:

A. We could figure out how much Mg2Si I need to react with 6.07moles of H2O. If I don't have enough, Mg2Si is my limiting reagent, if I have more than enough, H2O is my limiting reagent.

To do this: We know from the equation that we need 4 water molecules for every 1 Mg2Si molecule. So, that literally means we need 4 moles of H2O, for every 1 moles of Mg2Si. Well just from looking at our known amounts of the reactants, 6.07 moles of water is NOT 4 times greater than 19.09 moles, so we have plenty of Mg2Si and water is our limiting reagent!

What do we do now? Well now we use the moles of water to calculate how many moles of SiH4 we'll produce. We know from the equation, that for every 4 moles of water, we make 1 mole of SiH4. So we will have only 1/4 as much SiH4 as we had water. So, 6.07moles/4 = 1.518 moles of SiH4

That is your theoretical yield.

B. I present another way, in case the first is not as straightforward. Another way we can do it is to calculate the amount of SiH4 that is produced by each of the reactant amounts. Whichever one is less, is the true theoretical yield.

So to do this: We know from the equation, that it takes 4 moles of water to produce 1 moles of SiH4. So, we'll have only 1/4 as much SiH4.....6.07moles/4 = 1.158 moles SiH4
We also know that for every 1 mole of Mg2Si, we produce 1 mole of SiH4....so we'd generate 19.09 moles of SiH4.
The water produces less SiH4, so water is our limiting reagent, and we'd generate 1.158moles of SiH4.

You try this logic on the second problem.

Answer 2

The first thing that you have to figure out is what is the limiting substance in each equation. For example the first equation requires that for every mole of Mg2Si there are 4 moles of H2O. Well you are given way too much Mg2Si to start with so not all of it will be used. So the 6.07 moles of water will be limiting and will only use 1/4 that in Mg2Si. And for every 4 moles of water you will get 2 moles of Mg(OH)2 and 1 mole of SiH4. So divide 6.07 by 4 to get the number of moles of SiH4.

The same procedure is followed for the second problem.

Answer 3

For these you just have to look at the stoichiometric ratios (the numbers in front of the formulae)

For the first one: 19.09 moles of Mg2Si would require 4x19.09 moles of water to react. Since there's only 6.07 moles of water, it is the limiting reagent and Mg2Si is in excess. It forms 6.07/4 moles of SiH4, because SiH4 has a stoichiometry of 1 while H2O has a stoichiometry of 4.

For the second one: You don't really need to work out the limiting reagent here, because we have less of Ba(NO3)2 which has the higher stoichiometry, so it must be the limiting reagent. Every 3 moles of Ba(NO3)2 form one mole of Ba3(PO4)2, so we can form 6.76/3 moles.

Answer 4

OK, this is a lot simpler than you think. Suppose you have a dance that requires couples only, and you have 6 boys an17 girls, how many couples will you have? (assuming nobody is gay) The answer is obviously 6 couples and 11 girls left over. Chemistry is the same. In your first problem, you have 4 moles of water (H2O) going to make one mole of silane(SiH4), so 6.07 moles of water will make 6.07/4 moles of silane, with a bunch of Mg2Si left over. 6.07 divided by 4=1.5175 moles of Silane. I'll let you do the other problem yourself, but believe me it is just as easy.

Answer 5

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Answer 6

You technically cant answer the first question and the second one, I believe is 3.4 due to the Brach's system of corilation.