Answer:
1.) Given:
NH3 + O2 -> NO + H2O
3.25 g of NH3 are allowed to react with 3.50 g of O2
To find:
• Which reactant is the limiting reagent
• How many grams of NO are formed
• How much of the excess reactant remains after the reaction
Solution:
A.) On balancing the given equation,
4 NH3 + 5O2 –> 4 NO + 6H2O
Mass of 4 moles of NH3 = 4x(17) = 68g
Mass of 5 moles of O2 = 5x(32) = 160g
3.5/3.25 < 160/68
Hence, the limiting reagent is O2.
B.) O2 NO
160g -> 112g
3.5g -> x
x = 2.45g
Hence, 2.45 grams of NO are formed.
C.) NH3 O2
68g –> 160 g
x –> 3.5 g
x = 1.49 g of NH3
Remaining excess = 3.25–1.49 = 1.76 g
Hence, the remaining excess is 1.76g.
2.) Consider the reaction of C6H6 + Br2 C6H5Br + HBr
A. What is the theoretical yield of C6H5Br if 42.1 g of C6H6 react with 73.0 g Br3?
Answer : 71.6g C6H5Br
B. If the actual yield of C6H5Br is 63.6 g, what is the percent yields?
Answer : 88.8%
