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Problem #16: The reaction of 15.0 g C4H9OH, 22.4 g NaBr, and 32.7 g H2SO4 yields 17.1 g C4H9Br in the reaction below:
C4H9OH + NaBr + H2SO4 ---> C4H9Br + NaHSO4 + H2O
Determine:
(a) the theoretical yield of C4H9Br
(b) the actual percent yield of C4H9Br
c) the masses of leftover reactants, if any
Solution to a:
1) Determine the limiting reagent bewteen the first two reagents (the third reagent will be dealt with in step 2):
C4H9OH ⇒ 15.0 g / 74.122 g/mol = 0.202369 mol
NaBr ⇒ 22.4 g / 102.894 g/mol = 0.217700 molC4H9OH ⇒ 0.202369 /1 =
NaBr ⇒ 0.217700 / 1 =Between these two reactants, C4H9OH is limiting.
2) Compare C4H9OH to H2SO4 to determine which is limiting:
C4H9OH ⇒ 0.202369 mol
H2SO4 ⇒ 32.7 g / 98.0768 g/mol = 0.333412 molC4H9OH ⇒ 0.202369 /1 =
H2SO4 ⇒ 0.333412 / 1 =Between these two reactants, C4H9OH is limiting.
Overall, the above process shows that the limiting reagent for the entire reaction is C4H9OH.
3) Determine theoretical yield of C4H9Br:
There is a 1:1 molar ratio between C4H9OH and C4H9Br.This means 0.202369 mol of C4H9Br is produced.
0.202369 mol times 137.019 g/mol = 27.7 g (to three sig figs)
Solution to b:
17.1 g / 27.7 g = 61.7%
Solution to c:
1) Due to the 1:1 molar ratio:
0.202369 mol of NaBr is used up.
2) Therefore:
0.217700 minus 0.202369 = 0.015331 mol of NaBr remains.
The solution for sulfuric acid follows the same path as for NaBr. Conversion to grams is left to the reader.
Problem #17: Ozone (O3) reacts with nitric oxide (NO) dishcarged from jet planes to form oxygen gas and nitrogen dioxide. 0.740 g of ozone reacts with 0.670 g of nitric oxide. Determine the identity and quantity of the reactant supplied in excess.
Solution:
1) Wrte the balanced chemical equation:
NO + O3 ---> NO2 + O2
2) Calculate moles:
NO ⇒ 0.670 g / 30.006 g/mol = 0.0223289 mol
O3 ⇒ 0.740 g / 47.997 g/mol = 0.0154176 mol
3) Determine limitng reagent:
NO and O3 are in a 1:1 molar ratio. O3 is limiting, making NO the compound in excess
4) Determine quantity of excess reagent:
Based on the 1:1 ratio, we know 0.0154176 mol of NO is used upTherefore:
0.0223289 mol minus 0.0154176 mol = 0.0069113 mol of NO remaining
Quantity means grams:
0.0069113 mol times 30.006 g/mol = 0.207 g (to three significant figures)
Problem #18: and beyond will come someday!
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