I have one attempt on part A. Please help Iron(II) oxide reacts with carbon mono

Question

I have one attempt on part A. Please help

Iron(II) oxide reacts with carbon monoxide to produce iron and carbon dioxide. Fe03 (s) + 3C0(g)2Fe(s) + 3CO2 (g) Part A What is the percent yield of iron if the reaction of 65.4 g of ion(III oxide produces 15.6 g of iron? Express your answer with the appropriate units. HA The penent ywd of Fe .[ Value Units Submit My Answers Give Up Incorrect; One attempt remaining; Try Again Part B Express your answer with the appropriate units. ? The percent yield of CO2= Value Units

Explanation / Answer

The given reaction is --

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

Part A) From the reaction, we can see that 1 mole of iron (III) oxide produces 2 moles of Fe.

Now, the molar mass of Fe2O3 = 159.7 g/mol , molar mass of Fe = 55.85 g/mol

So, 159.7 g of iron (III) oxide produces = 2 x 55.85 g FE = 111.7 g Fe

Therefore, 65.4 g of iron (III) oxide will produce =( 111.7 / 159.7) x 65.4 g Fe = (7305.18 / 159.7) g Fe = 45.74 g Fe.

Now,

The percent yield of Fe = (actual mass / theoretical mass) x 100 %

=> % yield of Fe= (15.6 g / 45.74g) x 100 %

=> % yield of Fe = 0.341 x 100 %

=> % yield of Fe = 34.1 %

Part B) From the reaction we can see that 3 moles of carbon monoxide produces 3 mol of carbon dioxide. So, the molar ratio is 1:1

Now, the molar mass of CO(carbon monoxide) = 28.01 g and the molar mass of carbon dioxide(CO2) = 44.01 g/mol

So, 28.01 g CO produces = 44.01 g CO2

Therefore, 76.8 g CO will produce =( 44.01 /28.01) x 76.8 g CO2 = (3379.968 / 28.01) g CO2 = 120.67 g CO2

Now,

The percent yield of CO2 = (actual mass / theoretical mass) x 100 %

=> % yield of CO2= (84.3 g / 120.67 g) x 100 %

=> % yield of CO2 = 0.698 x 100 %

=> % yield of CO2 = 69.8 %

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