Iron(iii) oxide reacts with carbon monoxide to produce iron and carbon; fe2o3(s)+3co(> 2fe(s)+3co2(g). a) what is the percent yield for the reaction if 65.0 g of iron(iii) oxide produced 15.0 g of iron? b) what is the percent yield for the reaction if 75.0 g of carbon monoxide produces 85.0 g of carbon dioxide?

a) %yield= 33.00 %

b) %yield= 72.1 %

Explanation:

Step 1: Data given

Mass of iron(III) oxide = 65.0 grams

mass of iron produced = 15.0 grams

Molar mass of Fe2O3 = 159.69 g/mol

Molar mass of CO = 44.01 g/mol

Step 2: The balanced equation:

Fe2O3(s)+3CO(g) →2Fe(s)+3CO2(g)

Step 3: Calculate moles of Fe2O3

Moles Fe2O3 = 65.0 grams / 159.69 g/mol

Moles Fe2O3 = 0.407 moles

Step 4: Calculate moles Fe

For 1 mole Fe2O3 we'll have 2 moles Fe

For 0.407 moles Fe2O3 we'll have 2*0.407 = 0.814 molesFe

Step 5: Calculate mass Fe

Mass fe = 0.814*55.845 g/mol

Mass Fe = 45.46 grams = theoretical yield

Step 6: Calculate % yield

%yield = (actual yield/theoretical yield)*100%

%yield = (15.0/45.46)*100%

%yield= 33.00 %

b) What is the percent yield for the reaction if 75.0 g of carbon monoxide produces 85.0 g of carbon dioxide?

Step 1: Data given

Mass of CO = 75.0 grams

mass of CO2 produced = 85.0 grams

Molar mass of CO = 28.01  g/mol

Molar mass of CO2 = 44.01 g/mol

Step 2: The balanced equation:

Fe2O3(s)+3CO(g) →2Fe(s)+3CO2(g)

Step 3: Calculate moles of CO

Moles CO = 75.0 grams / 28.01 g/mol

Moles CO = 2.68 moles

Step 4: Calculate moles CO2

For 1 mole Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 2.68 moles CO we'll have 2.68 moles CO2

Step 5: Calculate mass CO2

Mass CO2= 2.68 * 44.01 g/mol

Mass CO2 = 117.95 grams = theoretical yield

Step 6: Calculate % yield

%yield = (actual yield/theoretical yield)*100%

%yield = (85.0/117.95)*100%

%yield= 72.1 %

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying