For each of the following reactions, 20.0 g of each reactant is present initially. Determine th

For each of the following reactions, 20.0 g of each...

Key Concepts

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, determining the maximum amount of product that can be formed. Identifying the limiting reactant is crucial because it governs the theoretical yield of the reaction.

Theoretical Yield

The theoretical yield is the amount of product that would be formed if the reaction proceeded perfectly, with every mole of the limiting reactant converted to product as predicted by the balanced equation. It is calculated based on stoichiometry and the initial quantities of reactants.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. Using the mole ratios from the balanced equation, one can convert between masses and moles to compute the amounts of substances consumed or produced.

Balanced Chemical Equations

A balanced chemical equation shows the relative number of moles of reactants and products involved in a chemical reaction, reflecting the conservation of mass and atoms. It is the foundation for stoichiometric calculations, as it provides the mole ratios necessary to determine how much product can be formed from given reactant amounts.

Molar Mass and Mass-to-Mole Conversions

Molar mass is the mass of one mole of a substance, and it allows the conversion between the mass of a substance and the number of moles. This conversion is critical in stoichiometry, as it helps in translating a given mass of a reactant to the number of moles that can react according to the balanced equation.

Transcript

00:01 So um it asks us, we have three different reactions here and we need to figure out which reacting is the limiting re agent.

00:08 We're given 20 g of all reactant in each chemical equation.

00:13 So let's start with the first.

00:15 So when i find limiting reagents i'm looking for which reaction is going to produce the least amount of product because that is my limiting reagent.

00:24 Um think of it like if i'm making sandwiches, it doesn't matter how much meat, cheese and bread i have, if i run out of bread and i have leftover meat and cheese, i still can't make sandwiches.

00:37 So it doesn't matter if we have um you know, one product is not used.

00:42 If the other product runs out, then that's then the reaction becomes to a halt and we're no longer allowed to make any more product.

00:49 So we need to figure out using some demento analysis and some multiple ratios.

00:56 How much can i produce with each reaction? how much product can produce, i mean, whatever one's the least amount is our limiting reagent.

01:05 So the first, the reason we have is aluminum some marks again, metallica.

01:14 And it's going to produce aluminum oxide.

01:20 And so um we're going to quickly figure out molar mass really quick for my product aluminum.

01:30 Alright, excuse me.

01:36 So all i'm going to do for all of these things when i try and find the molar mass of the compound, i'm going to look up on the periodic table of elements, the molar mass of each individual element that makes up my compound.

01:51 And so um you also need to take into account that subscript there, how many of each are in that compound? different aluminum.

02:00 I look it up on the periodic table of elements and it tells me that it's a 26 0.92, but i have two of them in this compound.

02:12 So i need to take that times two and then oxygen look it up, it tells me it's 15.999, but i have three.

02:21 So take that times three and then i'm gonna add it all together.

02:26 So 26.92 times two plus 15.999 times three.

02:33 And that's going to equal my molar mass.

02:35 So when i do all the multiplication, adding comes to a total of 101.961 grams per mole.

02:45 So that will be key anytime you have to find molar mass of a compound um you could look it up.

02:51 But another thing to do is you also have to look up on the prt developments the molar masses of all your different elements in that compound.

02:59 Um aluminum, that's another one that's on the periodic table, since it's an element we know that's gonna be 26.92 grams per mole oxygen is a little tricky because it's always diatonic.

03:13 So it's going to be um 15.999 times two.

03:19 Um so when we take 15.999 times two because there's a little subscript there, it's gonna be a 31 29 98 roll.

03:33 So now that we've got that taken care of um let's go ahead and start converting.

03:37 So if i have 20 g of each reacting, so i have 20 g of aluminum, i convert aluminum, two g.

03:48 Excuse me, two moles.

03:50 So for every one mole of aluminum, um i know that the cable told me i have 26.92 g.

03:59 And then my multiple ratio.

04:01 So that's what i need to look at for my balanced equation here, i get four moles for every two moles of product.

04:09 So i have four moles of aluminum, very two moles product.

04:19 And then for every one mole of product i get 101.961 g.

04:27 So in all these, i'm going to multiply all the numbers across the top.

04:32 I'll show you this the first time.

04:35 So you can see where i'm getting these numbers from.

04:38 But multiply all the numbers across the top.

04:41 And so when i multiply all the numbers across the top, i get 4,778.44.

04:50 And then i'm gonna take all the numbers across the bottom.

04:53 Okay, you can technically think this as a one.

04:55 So one times 26.92 times four times one.

05:01 So we multiply all the numbers across the bottom, i get a 107 .928.

05:11 So then i'm going to divide.

05:13 So now i'm going to take um 4078.44 divided by my 107.

05:21 And that's going to give me my answer for how much mass product that can make with 20 g of aluminum reaction.

05:30 So when i do the math plug into my calculator, i get roughly 37 0.8 g.

05:41 So now we see if i had 20 g of my oxygen.

05:47 Is that going to be less? if it's less than that tells me my oxygen as the limiting reagent.

05:52 If it's more than now, no, aluminum is my limiting reagent.

05:56 So again, same process, 20 g of oxygen.

06:01 I want to convert that to moles of oxygen.

06:04 And i know that since auctions i atomic, it's going to be 31.998 g.

06:10 Then i got to use my multiple ratio.

06:13 So i gotta look up what multiple ratio is from my balanced equation above.

06:23 Yeah, so it's a 3-2 molar ratio.

06:30 So three moles of oxygen for every two moles product.

06:35 So three moles oxygen.

06:38 Two more my product.

06:43 And then for every one mole of product, um i know that i get 101 .961g.

06:53 So when i multiply the number of the top and then divide by all the numbers multiplied by the bottom.

06:59 Um and that same process i did earlier.

07:02 I get 42.5 gramps.

07:06 This number is more.

07:08 So now i know my limiting reagent.

07:11 Is it going to be my aluminum? because once i run out of aluminum, i've only been able to make 37.8 g.

07:19 So i'm going to run out of aluminum first and then run out of oxygen.

07:24 So my aluminum for the second one let her be um we have some iron here and that's going to react with.

07:38 So to form some rust from iron oxide and hydrogen.

07:54 And um we're trying to find, it tells us in the question there.

07:59 Um in the parentheses, we're trying to find how much product can i make my um iron oxide, my iron three oxide.

08:09 So when we do this same process as before, we're just going to convert from 20 g of each product to or from 20 g of each reaction to product and whatever one makes the least amount is my limiting reagent.

08:23 So, um lower masks of iron, we look it up on the periodic table, tells us 55.845 easy molar mass of water.

08:42 I'm just going to add two hydrogen plus one oxygen on my periodic table.

08:48 And so when i add those all up, i get 18.015 g per mole.

08:54 And then same thing from my molar mass of my iron trioxide mhm...

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