For the equation Fe + 6HNO₃ ⇒ Fe(NO₃)₃ + 3NO₂ + 3H₂O 21.4 gram of HNO₃ is reacted with 20.3 grams of Fe.
a. What is the limiting reactant?
b. How many grams of Fe(NO₃)₃ will form?
c. Suppose only 4.02 grams of Fe(NO₃)₃ are recovered in the above reaction. What is the percent yield of Fe(NO₃)₃ will form?

Please show your work! Thank you!!

a. => LR is HNO₃

b. => 13.88 grams Fe(NO₃)₃

c. => 29.1%

Explanation:

a. Limiting Reactant:

Note => Determining LR is quite simple if you'll convert all data to moles and divide each mole value by the respective coefficient. The smaller value is the limiting reactant.

Given:   Fe      +                   6HNO₃ ⇒ Fe(NO₃)₃ + 3NO₂ + 3H₂O

          20.3g                          21.4g

         = 20.3g/56g/mole     =21.4g/63g/mole

         = 0.363 mole              = 0.340 mole

Divide by coefficient 1         Divide by coefficient 6

= 0.363/1 = 0.363                  = 0.340/6 = 0.057

Since 0.057 is the smaller value then HNO₃ is the limiting reactant. However, one must use the calculated mole values when applying to equation stoichometry.

b. Grams of Fe(NO₃)₃ Produced:

Problems like this can be greatly simplified if data is 1st converted to moles and solved with respect to equation reaction ratios. Then convert mole value to needed dimension.

Fe     +   6HNO₃ ⇒ Fe(NO₃)₃ + 3NO₂ + 3H₂O

Rxn Ratio between LR and Fe(NO₃)₃ => 1:6

0.340 mole HNO₃  =>  1/6(0.340mole Fe(NO₃)₃ Produced = 0.057mole Fe(NO₃)₃ x 242 g/mole = 13.8 grams of Fe(NO₃)₃ produced

c. Percent Yield:

Given:  Fe     +   6HNO₃ ⇒ Fe(NO₃)₃ + 3NO₂ + 3H₂O

Measured Lab Yield = 4.02 grams (given in problem)

Theoretical Yield = 13.8 grams (calculated in part b)

%Yield = (Lab Yield/Theoretical Yield)100%

= (4.02/13.8)100% = 29.1% (w:w)*

*w:w means calculation was based on weight values and not volume values.

the answer is c i took the test