Question

Worksheet 6 - Rev 10 Stoichiometry Exercise D - Stoichiometry 2 (5 points per answer) For this problem, identify the limiting reagent and calculate the grams of CO2 obtained in the reaction of 100.0 grams of C7H12OsN, with 100.0 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. C7H12O;N; + O2 CO2 + H2O + NH3 What is the limiting reagent? How many grams of CO2 will be produced? If 125 grams of CO, are produced, what is the % yield?

Answer 1

The reagent which is required in less amount to get the final product is known as limiting reagent.

A very easy way to understand the concept of limiting reagent is: observe the reaction very carefully, you will see that CO₂, H₂O and NH₃ all three of these products forms from C₇H₁₂O₅N₃, what that means is that let us assume all of C₇H₁₂O₅N₃ is consumed/reacted, now no matter how much O₂ you provide no product will form right?, so here the limiting reagent is C₇H₁₂O₅N_3, as it is required in less amount, or more crucial reagent.

all the other calculations are given in the following image:

Balanced equation of the reaction : 2 C9 HizOg Ng + 20 02 -> 14CO2 + 3H20 + 6NH, o Limiting Agent: Ca H₂2 OG N3 (Answer) Molar mass of Cg Miz Og Ng = (7X12) + (12 xI) + (5x16)+ (3x14) = 218 g/mol Molar mass of co2 = 44g/mol from the equation we can see that: from (2x2189) Ca Hiz Of Nz produces (14x44g of co2 = 616g = 4369 6/6 436 .: 1009 C7 HIZ 05 Ng will produce = x100 g CO2 = 141.28 g CO2 (Auswer → Theoritical yield = 141.289 Actual yield = 125g .. 7. Yield = 2 x 100%. = 88.48%. (Answers 1.28