Calculate the theoretical yield of product (in moles) for each
i...

Question

Calculate the theoretical yield of product (in moles) for each 
initial amount of reactants. 
3 Mn(s) + 2 O2( g)¡Mn3O4(s) 
b. 4 mol Mn, 7 mol O2

Accepted Answer

Welcome back everyone. We need to calculate the theoretical yield in moles, giving the starting amounts of nine moles of lead And 19 moles of bro. Mean for our given reaction below. So according to our reaction, we have one mole of solid lead reacting with two moles of liquid bromine to form our solid lead for bromide as a product. And that's just one mole formed. So we want to see whether this is balanced and analyzing each atom in our compounds here, we can see that on both sides of our equation. We have a balanced amount of each type of atom and so we can confirm that this is a balanced reaction which is always important to do first. And because we need to find theoretical yield and moles, we wanna therefore find our limiting reactant to determine our theoretical yield of how much product we will actually produce. So using the info from the prompt for our first reactant, we have nine moles of lead for our given reaction. And so we're going to use that to start off with our stock geometry. So we have nine moles of lead and we're going to multiply by a conversion factor where we want to cancel out moles of lead to find moles of our product that is produced. And so we can see from our formula that we can use a multiple ratio. So we have one mole of lead according to our formula which produces one mole of our product. And so we would plug that in, We have one mole of lead in the denominator for one mole of our lead for bromide. And we would see that we cannot cancel out moles of lead, which leaves us with moles of our product lead for bromide that is produced. And this gives us a value of nine moles of lead for bromide. That should be produced from our nine moles of our first reactant. And sorry, this should be a subscript here. So we want to go ahead and compare this to our molds of our second reactant Remain here, which according to the prompt, we have 19 moles of so we'll use the color red here. Or we'll use black. Actually we have 19 moles of bromine and we're going to multiply this by the multiple ratio from our balanced equation where we see we have two moles of liquid roaming, which produces one mole of our product lead for bromine. So plugging our two moles of roaming and denominator, we see that according to our equation we produce one mole of our lead for bromide. And so we can cancel out our moles of roaming And what we will produce is a total of 19 or sorry, since it's divided by two, we would produce 9.5 moles of our lead for bromide and we would see that based on these two amounts of our product produced. We want to recall that the lower amount is going to be or determine our limiting reactant. And because we see that we have a lower amount of our product lead for bromide produced from the reactant, nine moles of lead. We would say that therefore lead is our limiting reactant and our theoretical yield. So we're using T Y for short for theoretical yield is going to equal nine moles of our lead for bromide that is produced. And so for our final answer, that is going to correspond to choice D in the multiple choice where we have nine moles of our lead for bromide produced. So what's highlighted in yellow represents our final answer in our solution. I hope that everything that I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Calculate the theoretical yield of product (in moles) for each initial amount of reactants. 3 Mn(s) + 2 O2( g)¡Mn3O4(s) b. 4 mol Mn, 7 mol O2

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