For this problem, identify the limiting reagent and calculate the grams of CO2 o

Question

For this problem, identify the limiting reagent and calculate the grams of CO2 obtained in the reaction of 150.0 grams of C7H12O5N3 with 150.0 grams of oxygen. If 125 grams of CO2 is actually produced, what is the % yield. The equations are not balanced. They were balanced in Exercise B. Use those coefficients to do these calculations.

C7H12O5N3 + O2 CO2 + H2O + NH3

What is the limiting reagent? ____________________

How many grams of CO2 will be produced? ____________________

If 125 grams of CO2 are produced, what is the % yield? ____________________

Explanation / Answer

4C7H12O5N3 + 21O2 28CO2 + 6H2O + 12 NH3

no of moles of C7H12O5N3    = W/G.M.Wt

                                             = 150/218    = 0.68 moles

no of moles of O2    = W/G.M.Wt

                               = 150/32   = 4.6875moles

21 moles of O2 react with 4 moles of C7H12 O5N3

4.6875 moles of O2 react with = 4*4.6875/21    = 0.893 moles C7H12 O5N3 is required

C7H12 O5N3 is limiting reagent

4 moles of C7H12O5N3 react with O2 to gives 28 moles of CCO2

0.68 moles of C7H12 O5N3 react with O2 to gives = 28*0.68/4   = 4.76moles of CO2

mass of CO2 = no of moles * gram molar mass

                     = 4.76*44   = 209.44g

percnt yiled    = actual yiled*100/Theoretical yield

                     = 125*100/209.44    = 59.7% >>>>answer

C7H12O5N3

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