Limit Reagent, Chem Question 2 SO2 (g) + O2 (g) → 2 SO3 (g) If 25.0 g of SO2 react, how many grams of SO3 could be produced?
Use the balanced equation to answer a-e
a. If 25.0 g of SO2 react, how many grams of SO3 could be produced?
b. If 25.0 g of O2 react, how many grams of SO3 could be produced?
c. What is the limiting reagent, based on answers of a and b?
d. What is the theoretical yield (in grams) of SO3?
e. If the reaction done using 25.0 g of each reactant produced only 10.8 g of SO3, what is the percent yield?
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a. 25.0 g SO2 x 1mole SO2/64.0 g x 2 moles SO3/ 2 moles SO2 = 0.391 moles SO3
0.391 moles SO3 x 80.0 g/1mole SO3 = 31.2 g SO3
b. 25.0 g O2 x 1mole O2/32 g x 2 moles SO3/1mole O2 = 1.56 moles SO3
1.56 moles of SO3 x 80.0 g / 1mole SO3 = 125 g SO3
c. One cannot obtain more than 31.2 g SO3, so SO2 is the limiting reactant.
d. 31.2 grams SO3 is the theoretical yield
e. Percent yield = actual amount obtained/ theoretical * 100
10.8 grams / 31.2 g * 100 = 34.6%