Imagine you are working on ways to improve the process by which Iron ore containing Fe2O3 is converted into Iron: Fe2O3(s) + CO(

Question

3 years ago

5

Imagine you are working on ways to improve the process by which Iron ore containing Fe2O3 is converted into Iron: Fe2O3(s) + CO(

g) Fe (s) + CO2(g) If you start with 150g of Fe2O3 as the limiting reactant, what is the theoretical yield of Fe? If your actual yield is 87.9g, what is the percent yield?

Chemistry

1 answer:

Goryan [66] · Answer 1 · 2022-06-10T05:41:48+03:00

Answer:

1. theoretical yield of Fe = 105g.

2. Percentage yield of Fe = 83.71%

Explanation:

Step 1:

Data obtained from the question. This includes:

The equation for the reaction:

Fe2O3(s) + CO(g) -> Fe(s) + CO2(g)

Mass of Fe2O3 = 150g

Actual yield of Fe = 87.9g

Theoretical yield of Fe =?

Percentage yield of Fe =.?

Step 2:

Balancing the equation.

Fe2O3(s) + CO(g) -> Fe(s) + CO2(g)

The above equation can be balanced as follow:

There are 2 atoms of Fe on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Fe as shown below:

Fe2O3(s) + CO(g) -> 2Fe(s) + CO2(g)

There are a total of 4 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of CO and 3 in front of CO2 as shown below:

Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)

Now the equation is balanced.

Step 3:

Determination of the mass of Fe2O3 that reacted and the mass of Fe produced from the balanced equation. This is illustrated below:

Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)

Molar Mass of Fe2O3 = (56x2) + (16x3) = 112 + 48 = 160g/mol

Molar Mass of Fe = 56g/mol

Mass of Fe from the balanced equation = 2 x 56 = 112g

From the balanced equation,

160g of Fe2O3 reacted.

112g of Fe is produced.

Step 4: