5. In an experiment, 169 g of FeCr2O4, 298 g of K2CO3, and...

5. In an experiment, 169 g of FeCr2O4, 298 g of K2CO3, and 75.0 g of O2 were reacted. If the amount of K2CrO4 isolated was 185 g, calculate the percent yield of the reaction. Also, calculate the amount in grams of excess reagents, if any. Molar mass: 223.84 g/mol 138.21g/mol 31.998g/mol 194.19g/mol 4FeCr2O4(s) + 8K2CO3(s) + 7O2(g) -> 8K2CrO4(s) + 2Fe2O3(s) + 8CO2(g)

Transcript

00:03 So we're given an experiment that involves three reactants.

00:06 We have to calculate the percentage yield as well as the mass of the excess reagent.

00:12 So let's start by writing the equation of the reaction.

00:15 4 moles of fecr2o4 plus 8 moles of k2co3 plus 7 moles of oxygen gas will give you 8 moles of k2cro4 plus 8 moles of co2 and plus 7 moles of fe2o3.

01:00 So we're going to pick each one and try to find the mass of the product obtained.

01:07 The limiting reagent will be the one that gives us the smallest amount of the product.

01:13 And that amount of product will also be the theoretical yield so let's start with the first one which is fecr2o4o6 not sure i can see it clearly so we have 169 grams of fecr2o4 we're going to change that to mole so that's one 1 mole of fecr2o4 and the molar mass is 223 .84 grams of fecr2o4.

01:58 We change that to mole of the product which is this.

02:03 This is 1 to 2.

02:05 So you have two moles of this, k2cro4 and one mole of fecr2o4.

02:22 And then we change it to mass.

02:24 The mass of the product, molar mass of the product is 194 .19 gram of k2cr2 and then this is one mole of k2.

02:39 Let's look at this.

02:42 This and this will cancel out.

02:43 All of this will cancel out.

02:49 So the mass of k2cro4 would be 169 multiplied by 2 multiplied by 194 .19 divided by 223 .84.

03:06 And this would give you 293 .2 grams.

03:14 That's it.

03:16 Now let's look to the others.

03:18 The next one is k2co3.

03:22 We're given 298 grams of that.

03:24 298 grams of k2co3.

03:28 We'll first change it to mole.

03:30 One mole of k2co3 has a mass of 138 .21 grams of k2co3.

03:42 Then the mold right one to one the ratio about 8801 mold of k2cr04 and one mold of k2co3 then we change to mass 194 .19 grams of k2cr04 and one mold of k2cr04 so let's look at the mass this and this will go and this will go this and this will go so the mass of k2 cro4 that will be produced will be 298 multiplied by 194 .19 and we'll divide that by 138 .21 and this would give us 418 .7 grams next, we will have oxygen.

04:46 We are told that oxygen, the mass was 75 grams, so 75 .0 grams of oxygen.

04:55 Change it to 1 mole of oxygen, that has a mass of 31 .998 grams oxygen...

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