An Giang info: Youtube daily report Dec 2 2018

This is part three of the stoichiometry 101 series. In this video we're gonna

look at solving mass A to mass B question. Here's the stoichiometric plan we'll be

using. I've explained the plan in the first two

videos so do check them out if you haven't seen them. I go through in detail

how the plan is developed and how to use it to set up solutions for

stoichiometric problems. The link is on the top and in the description box below.

Now, let's read the question. We're given an equation and asked to find the moles

of Na2S2O3 that are needed to react with 42.7 grams of AgBr. So the

first thing we need to do is find out what we have and that's 42.7 g

of AgBr and what does the question wants? It's moles of Na2S2O3.

Since we're dealing with two substances here, which is AgBr and

Na2S2O3, we'll call them A and B respectively. A is for what's given. B is

for what we need to find. So our starting point is what we have, and ending point is

what the question wants. Next, we'll tag our start and end point on our stoichiometry

plan. So our start point is grams A and end point is mole B. Looking at the

plan, it will take two steps to go from grams A to moles B. Before we begin setting

up to solve the question we need to make sure that we have a balanced equation. so,

if you need a refresher on balancing equation, I'll link a video on the top.

So, it looks like we can easily balance this equation by placing a 2 in front of

Na2S2O3. With that we are now ready to set up the solution.

In step 1, we'll go from grams A to moles A and in

step two we'll go from mole A to mole B. A in this case, like I mentioned before is

AgBr and B is Na2S2O3. Step 1 is to divide the given mass, 42.7 grams with

the molar mass of AgBr. That comes out to be 187.77 g/mol.

If you need a refresher on calculating molar mass, I'll link the

video on the top as well. So we take 42.7 grams and divide it

with 187.77 g/mol and we get 0.2274 mol.

What we have just calculated is mole A, which is the mole

of AgBr. Moving on to step two, we are supposed to multiply what we got from

step one with the coefficient ratio from the balanced equation. Coefficient B is

the number in front of Na2S2O3, which is 2. Coefficient a is the number

in front of AgBr, which is 1, so, we take 0.2274 mol and multiply with 2

over 1 and that gives us 0.4548 mol. If you need to present it in three

significant figures, then it's 0.455 mol. That's the mole of

Na2S2O3,3 which is what the questions wants us to find.

To recap, we start off by reading the question and figuring out the

information we have and what the question wants. Then, we figure out the

pathway using the plan. Before we proceed, if the question involves two substances,

A and B then we need to make sure that our equation is balanced. Once we get

that sorted out we follow the set up and solve the problem. You can solve

practically any stoichiometric questions this way successfully every time.

For more examples, watch out for more videos in this stoichiometry 101 series.

Hope the video was helpful. Do subscribe and thanks for watching!

For more infomation >> Stoichiometry problem grams to mole | Stoichiometry 101: Part 3 - Dr K - Duration: 4:08.

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Stoichiometry problem grams to mole | Stoichiometry 101: Part 3 - Dr K - Duration: 4:08.