1. suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theoretical yield of the product?

2. if 0.8922 g of magnesium oxide is obtained from the reaction indicated in #1 above, what would be the percent yield of the magnesium oxide?

3. suppose the percent yield was calculated to be over 100%, what possible reasons could you give to account for it?

Answer :

(1) The theoretical yield of product, is, 1.257 grams.

(2) The percent yield of is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

Now we have to calculate the moles of

The balanced chemical reaction is,

From the reaction, we conclude that

As, 2 mole of react to give 2 mole of

So, 0.03142 moles of react to give 0.03142 moles of

Now we have to calculate the mass of

Theoretical yield of = 1.257 g

Experimental yield of = 0.8922 g

Now we have to calculate the percent yield of

Therefore, the percent yield of is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

0.88m hope i am correct : )

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o