How many grams of each product result from the following reactions, and how many grams of which reactant is left over? (b)

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Hello Everyone in this video, we're trying to identify which reagent is going to be in excess. And then with that known, we can go ahead and calculate for the remaining mass in grams that's basically not used. All right. So let's go ahead and first read out the balanced equation such as our baseline equation first is of course our H two S reacting with C A B R two and that yields H B R and c A S. So you see, first of all, right off the bat that on the left side we have two atoms of hydrogen and I'll just wear this out but two atoms of growing. But on the right side we only have one of each. We simply add a coefficient of two. And now this equation is our balanced chemical equation. Alright, so I'm gonna go ahead and actually list out all the molar masses of our reagents and this of course can either be found online or given to you by a professor or even if you saw this on your own. Alright, so for H two S we have 34.8 to of course all these units will be in grams per mole. All right. And then for RCABR two, that is 199.886 for our HBR. That is 80.912 and last CCS. That is equal to 72.144. Now solving for the moles of reactant to go ahead and hopefully find a access and limiting reagent. So for our moles of reactant. So our starting materials or given 1.6 g of H to us, we want to convert this into molds when we use the molar mass. So for every one mole we have 34.82 g. I'm putting this in your calculator. We get the value of 0.0464 course moles of H two s. Alright, so for our second surgery, agents have the 9.5 g of C A B R two. Using the molar mass. Again, to convert into moles we have for every one mole to give us 99.886 g. This will give us a value of 0. Moles of CABR two. So looking at our balanced equation here, they were starting regions will have a 1 to 1 ratio. Therefore we will need to take that into account. We just look simply at the moles of our reactant and find our limiting reagent. So you can see here that the moles of H two S is going to be smaller. Therefore this will be our limiting reagent. Now we can go ahead and solve for the mass of the products with us produced of course, knowing the information now that our H- two s is our limited regions. So again, we're selling now for the mass of products as produced. So starting off with my calculations using my lemony reagent, we have zero moles of H two S. We use historical metric motor mole ratio. So we have for every one mole of H two S we get two moles of H B. R. We see now that the map the moles of H two S cancels. Now we have the moles of H B. R. We want the units being grams. So of course we use the moller mass. Okay, so for every one mole of H B. R, we have 80.912 g of H B. R. And that would give us the value of 7.6 g of H. B. R. All right now, the mask for my second product here. So again, starting off with my limiting reagent of 0. moles of H two S. Just repeating the same process. We're using the historic parametric multiple ratio. So for every one mole of H two S, we have one mole of C. A. S. Using the molar mass. That for every one mole of CS we have 1 or rather 72.144 g of R. C. S. So once we put that into a calculator we can get the value of 3.4 and four units. Let's see the moles of H two S. Councils and the moles of C. S. Councils. So the units will be grass of C A. S. So that's the mass of the products that's produced. So that's going to be whatever answers. Alright, now calculate for the mass our excess reactant that remains. So let's first solve for how much of CBR two is actually used first. So the mass of CABR two used. So similar to finding the mass of the products. We're starting off our calculations of our lemony regent. So 0.0469 moles of H two S. Again using the multiple ratio. Of course. So for every one mole of H to us with one mole of C a B R two and then using the molar mass. Then for every one mole of C a B R two we have 1 99. g of C A B R two. You see now the most HTS cancels and molds of CBR two cancels. Now putting all these numerical values into my calculator, I'll get the numerical value to be 9.4 units being grams of C a B R two. Okay, now that we know how much of that reacting we actually be using, we now sell for how much is being our remains. So the mass of CABR two that remains are basically unused. Well we're starting off with 9.5 g and we have said to use 9.4 g. That gives us the value of 0.1 g. Therefore the mass of C r b r two is 0.1 g. Our last and final answer for this problem.

How many grams of each product result from the following reactions, and how many grams of which reactant is left over? (b)

Verified Solution

Key Concepts

Stoichiometry

Limiting Reactant

Molar Mass