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

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Hello. Everyone in this video, we're given this chemical reaction right here. We're doing in this problem is to calculate the amount of each product in grams as well as to identify which reaction has been access. And to calculate its mass in grams. So first we're gonna go ahead and realize that we need our molar masses. So we'll just find the molar mass of each of our agents. So I'm just using values that I have but you can find this either in your textbook or even given to you by a professor. So for the first reagent, which is NH four br the molar mass of this is 97.946 units being grams per mole Next for AGCL four, That's equal to 207.35. Then for a product side we have NH four cl four That has a molar mass of 117.492. Their last product is going to be a g B R. This has a molar mass of 187.804. Now, let's go ahead and use these to calculate the molds of each reactant that we have. So the products on the left side, N H four, B R and a g c L 04. So let's go ahead and start off with the first one. Then For NH four BR. So given in the chemical reaction that we're starting off with the .8 g of this. Then we'll go ahead and use the molar mass to get our units to be in moles. So the denominator for our conversion factor will be 97.946. And then top will just be moles. So you can see that the grams unit will cancel. So putting this into my calculator, I get the value of 0.0388. Most of our agents to doing the exact same thing for our second starting material. So we're doing it for AGCL 04. So given to us in the equation, we're starting off with 6.3 g. Then again using the molar mass. So denominator will be grams. That's 2 07.35. And again, top will just be one more. You can see that the grams unit will cancel Putting this into my calculator. I get the value of 0.0304 moles. And because our whole reaction is a 1-1 ratio, we can go ahead and just use these products here to find our limiting reagent. We can see that we're comparing to just these two values here. AGCL- 04 has the less amount of moles. So are limiting reagent will be this right here. Let's go ahead and highlight that. But this is our limiting regions. Alright, now that we have that information. We can go ahead and calculate for the mass of the products that are produced. So let's just scroll down to have more space. All right, so X is equal to the mass of n H four C L 04 whenever products. So again, we're starting off with the Limiting reagent here and that's zero molds of a G cell for then we'll do a multiple conversion. So we have one more Of the AGCL- 04 for every one mole of r NH four cl oh four. Again, that's because we have a 1 to 1 mole ratio. And then now to convert this into grams, we can go ahead and use the molar mass. So for every one mole of this product, We have 117.49 g of the product. And for unit cancelation, we can see that the moles will cancel leaving us with just the grounds. So once we put everything into the calculator, we see that X is equal to 3.6 grams of this product. That's one of the answers for a problem. So go ahead and highlight that. Next we can go ahead and calculate for why? So why is equal to the mass of our second product? That's a g B R. Again, I'm going to go ahead and start my dimension analysis with our limiting reagent that 0.0304 Moles of AGCL four. Again, I'll do a multi mo conversion. So for every one mole of our lemony regions, We get one mole of our product that would go ahead and use the molar mass to convert our moles into grams. So for every one mole of our products, We have a 187.804g of a g B R. Again, for our units we can see that the molds will cancel leaving us with just the units of grams. So putting this into the calculator, I get the value of 5.7 g of a G B R. So again, why then is equal to 5.7 g of AGBR. So that's our second answer for this question again scrolling. So now calculate for the mass of our NH four br which is our access. We're just gonna go ahead and start the dimensional analysis again with our limiting reagent. So that's 0.0304 units being mold of AGCL four, I'll go ahead and do a multiple conversion. So for every one mole of our limited regions we get one mole of our access. So NH four B r. Alright then of course we're gonna go ahead and use the fuller mass to convert our mole unit into grams. So one mole of NH four br gives us 97.946 g of NH four Br So you see now that the moles will cancel leaving us with just the units of grams. So putting this into my calculator, I get the value of 3.0 g of NH four B r so the mass of our access to NH four br that remains Is equal to when we subtract the initial amount 3.8g with 3.0 And that gives us a difference of 0.8 graphs so that right, there is going to be my 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? (a)

Verified Solution

Key Concepts

Stoichiometry

Limiting Reactant

Molar Mass