| Limiting Reagent Problems #1-10 | Limiting reagent tutorial | Stoichiometry Menu |
Problem #11: The equation for the reduction of iron ore in a blast furnace is given below. (a) How many kilograms of iron can be produced by the reaction of 7.00 kg of Fe2O3 and 3.00 kg of CO? (b) How many kilograms of the excess reagent remains after reaction has ceased?
Fe2O3 + 3CO ---> 2Fe + 3CO2
Solution to a:
1) Determine the limiting reagent:
Fe2O3 ---> 7000 g / 159.687 g/mol = 43.836 mol
CO ---> 3000 g / 28.01 g/mol = 107.105 molFe2O3 ---> 43.836 mol / 1 mol = 43.836
CO ---> 107.105 mol / 3 mol = 35.702CO is the limiting reagent.
2) Use the CO : Fe molar ratio:
3 is to 2 as 107.105 mol is to xx = 71.403 mol of Fe produced
3) Convert to kilograms of Fe:
(71.403 mol) (55.845 g/mol) = 3987.52 gto three sig figs this is 3.99 kg of iron
Solution to b:
1) Use Fe2O3 to CO molar ratio
1 is to 3 as x is to 107.105 molx = 35.702 mol of Fe2O3 consumed
2) Determine mass remaining:
(35.703 mol) (159.687 g/mol) = 5701 g consumed7000 g − 5701 g = 1299 g
to three sig figs, this is 1.30 kg
Problem #12: The reaction of 4.25 g of Cl2 with 2.20 g of P4 produces 4.28 g of PCl5. What is the percent yield?
Solution:
1) First, a balanced chemical equation:
P4 + 10Cl2 ---> 4PCl5
2) Get moles, then limiting reagent:
P4 ---> 2.20 g / 123.896 g/mol = 0.0177568 mol
Cl2 ---> 4.25 g / 70.906 g/mol = 0.0599385 molP4 ---> 0.0177568 / 1 = 0.0177568
Cl2 ---> 0.0599385 / 10 = 0.00599385Cl2 is the limiting reagent.
3) How many grams of PCl5 are produced?
Use Cl2 : PCl5 molar ratio of 10 : 4 (or 5 : 2, if you prefer)5 is to 2 as 0.0599385 is to x
x = 0.0239754 mol of PCl5 produced
(0.0239754 mol) (208.239 g/mol) = 4.99 g ( to three sig figs)
4) Determine percent yield:
(4.28 g / 4.99 g) x 100 = 85.8%
Notice how asking about percent yield (oh, so innocuous!) forces you to go through an entire limiting reagent calculation first.
Problem #13: 35.5 g SiO2 and 66.5 g of HF react to yield 45.8 g H2SiF6 in the folowing equation:
SiO2(s) + 6 HF(aq) ---> H2SiF6(aq) + 2 H2O(l)
(a) How much mass of the excess reactant remains after reaction ceases?
(b) What is the theoretical yield of H2SiF6 in grams?
(c) What is the percent yield?
Solution to (a):
1) Must determine limiting reagent first (even is it not asked for in the question):
SiO2 ---> 35.5 g / 60.084 g/mol = 0.59084 mol
HF ---> 66.5 g / 20.0059 g/mol = 3.324 molSiO2 ---> 0.59084 mol / 1 mol = 0.59
HF ---> 3.324 mol / 6 mol = 0.554HF is limiting.
2) Determine how much SiO2 remains:
The SiO2 : HF molar ratio is 1 : 61 is to 6 as x is to 3.324 mol
x = 0.554 mol of SiO2 used up
0.59084 mol − 0.554 mol = 0.03684 mol of SiO2 remains
(0.03684 mol) (60.084 g/mol) = 2.21 g (to three sig figs)
Solution to (b):
There are 0.59084 mol of SiO2SiO2 : H2SiF6 molar ratio is 1 : 1
therefore, 0.59084 mol of H2SiF6 produced
(0.59084 mol) (144.0898 g/mol) = 85.1 g (to three sig figs)
Solution to (c):
(45.8 g / 85.1 g) (100) = 53.8%
Problem #14: Gaseous ethane reacts with gaseous dioxygen to produce gaseous carbon dioxide and gaseous water.
(a) Suppose a chemist mixes 13.8 g of ethane and 45.8 g of dioxygen. Calculate the theoretical yield of water.(b) Suppose the reaction actually produces 14.2 grams of water . Calculate the percent yield of water.
Solution to (a):
1) Write the balanced equation:
2C2H6 + 7O2 ---> 4CO2 + 6H2O
2) Determine limiting reagent:
C2H6 ---> 13.8 g / 30.0694 g/mol = 0.45894 mol
O2 ---> 45.8 g / 31.9988 g/mol = 1.4313 molC2H6 ---> 0.45894 / 2 = 0.22947
O2 ---> 1.4313 / 7 = 0.20447Oxygen is limiting.
3) Determine theoretical yield of water:
The oxygen : water molar ratio is 7 : 67 is to 6 as 1.4313 mol is to x
x = 1.2268286 mol of water
4) Convert moles of water to grams:
(1.2268286 mol) (18.015 g/mol) = 22.1 g (to three sig figs)
Solution to (b):
(14.2 g / 22.1 g) (100) = 64.2%
Problem #15: A 0.972-g sample of a CaCl2 ⋅ 2H2O and K2C2O4 ⋅ H2O solid salt mixture is dissolved in 150 mL of deionized water, previously adjusted to a pH that is basic. The precipitate, after having been filtered and air-dried, has a mass of 0.375 g. The limiting reactant in the salt mixture was later determined to be CaCl2 ⋅ 2H2O
(a) How many grams of the excess reactant, K2C2O4 ⋅ H2O, reacted in the mixture?
(b) What is the percent by mass of CaCl2 ⋅ 2H2O?
(c) How many grams of the K2C2O4 ⋅ H2O in the salt mixture remain unaffected?
Solution to (a):
1) Write the balanced chemical reaction:
CaCl2 + K2C2O4 ---> CaC2O4 + 2KCl
2) Determine moles of calcium oxalate that precipitated:
0.375 g / 128.096 g/mol = 0.0029275 mol
3) Determine moles, then grams of potassium oxalate:
The K2C2O4 : CaC2O4 mole ratio is 1:1therefore, 0.0029275 mol of potassium oxalate monohydrate reacted
0.0029275 mol) (184.229 g/mol) = 0.53933 g
To three sig figs, 0.539 g
Solution to (b):
1) Determine moles, then grams of calcium chloride that reacted:
The CaCl2 : CaC2O4 mole ratio is 1:1therefore, 0.0029275 mol of calcium chloride dihydrate reacted
(0.0029275 mol) (147.0136 g/mol) = 0.43038 g
To three sig figs, 0.430 g
2) Determine mass percent of calcium chloride:
(0.430 g / 0.972 g)(100) = 44.24%
Solution to (c):
1) Determine total mass that reacted:
0.430 g + 0.539 g = 0.969 g
2) Determine mass of excess reactant that remains:
0.972 g − 0.969 g = 0.003 gRemember, we were given only the total mass of the mixture. Also, we know the remains of the mixture contain zero calcium chloride, since it has been determined to be the limiting reagent.
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 = 0.202369
H2SO4 ---> 0.333412 / 1 = 0.333412Between 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) (137.019 g/mol) = 27.7 g (to three sig figs)
Solution to (b):
(17.1 g / 27.7 g) (100) = 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 − 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 up. Therefore:0.0223289 mol − 0.0154176 mol = 0.0069113 mol of NO remainingQuantity means grams:
(0.0069113 mol) (30.006 g/mol) = 0.207 g (to three significant figures)
Problem #18: If 1.24 g of P4 reacts with 0.12 g of H2, to give 1.25 g of PH3, determine percent yield.
Solution:
1) First, the balanced equation:
1⁄4P4 + 3⁄2H2 ---> PH3Decided to do it with fractions.
2) Determine moles of P4 and H2:
1.24 g / 123.896 g/mol = 0.01001 mol
0.12 g / 2.016 g/mol = 0.059524 mol
3) Determine the limiting reagent:
0.01001 / 0.25 = 0.04004
0.059524 / 1.5 = 0.039683H2, by a nose!
4) Determine moles of PH3 that can be made from 0.059524 mol of H2:
The molar ratio is 1.5 to 11.5 is to 1 as 0.059524 mol is to x
x = 0.039683 mol
5) Determine mass of PH3 (this would be the 100% yield amount):
(0.039683 mol) (33.9977 g/mol) = 1.35 g (to three sig figs)
6) Percent yield:
(1.25 / 1.35) * 100 = 92.6%
Problem #19: What mass of phosphorus (P4) is produced when 41.5 g of calcium phosphate, 26.5 g of silicon diioxide, and 7.80 g of carbon are reacted according to the eqation:
2Ca3(PO4)2 + 6SiO2 + 10C ---> P4 + 6CaSiO3 + 10CO
Solution:
1) Calculate the moles of each reactant:
Ca3(PO4)2 ---> 41.5 g / 310.174 g/mol = 0.133796 mol
SiO2 ---> 26.5 g / 60.084 g/mol = 0.44105 mol
carbon ---> 7.80 g / 12.011 g/mol = 0.649405 mol
2) We now write, for each reactant, the ratio of moles present to the coefficient:
Ca3(PO4)2 ---> 0.133796 / 2 = 0.066898
SiO2 ---> 0.44105 / 6 = 0.07351
carbon ---> 0.649405 / 10 = 0.0649405 <--- smallest value
3) Carbon is the limiting reagent.
4) Determine moles of P4 produced:
The C to P4 molar ratio is 10 to 110 is to 1 as 0.649405 is to x
x = 0.0649405 mol of P4
5) Determine mass of P4 produced:
(0.0649405 mol) (123.896 g/mo) = 8.045868 gFollowing the rule for rounding with five, 8.04 g.
Problem #20: 320 g of sulfur dioxide, 32 g of oxygen, and 64 g water react. (a) which reactant is the limiting reagent and (b) how much H2SO4 is produced?
2SO2 + O2 + 2H2O ---> 2H2SO4
Comment: I'm going to solve this problem by doing three mass-mass calculations using SO2, then O2, then H2O.The answer that yields the smallest amount of H2SO4 will answer both (a) and (b).
Solution using SO2:
determine moles SO2 ---> 320 g / 64 g/mol = 5.0 moluse 1:1 molar ratio to determine 5.0 mol of H2SO4 produced
determine mass of H2SO4 ---> (5.0 mol) (98 g/mol) = 490 g
Solution using O2:
determine moles O2 ---> 32 g / 32 g/mol = 1.0 moluse 1:2 molar ratio to determine 2.0 mol of H2SO4 produced
determine mass of H2SO4 ---> (2.0 mol) (98 g/mol) = 196 g
Solution using H2O:
determine moles H2O ---> 64 g / 18 g/mol = 3.6 moluse 1:1 molar ratio to determine 3.6 mol of H2SO4 produced
determine mass of H2SO4 ---> (3.6 mol) (98 g/mol) = 353 g
The final answer is that O2 is the limiting reagent and that 196 g of H2SO4 is produced.
I copied this problem from an "answers" website. The answerer has a nice explanation on how to determine the limiting reagent. Here it is:
Think of the coefficients as a "recipe" for the reaction. The "recipe" calls for 2 mol SO2, 1 mol O2, and 2 mol H2O. Convert each of the given masses into moles and find out how many "recipes" each substance can make. The substance that makes the fewest "recipes" is the limiting reagent.
One "recipe" is:
2 mol SO2 = 2 x 64.1 g = 128 g
1 mol O2 = 32.0 g
2 mol H2O = 2 x 18.0 = 36 g
Determine "recipes:"
320 g SO2 x (1 recipe / 128 g) = 2.50 recipes
32 g O2 x (1 recipe / 32.0 g ) = 1.00 recipes
64 g H2O x (1 recipe / 36 g) = 1.78 recipes32 g O2 is the limiting reagent because it makes the fewest "recipes." You would use the 32 g O2 to find the amount of H2SO4 produced.
Problem #21: Consder the following reaction:
C4H9OH + NaBr + H2SO4 ---> C4H9Br + NaHSO4 + H2O
If 15.0 g of C4H9OH react with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction?
Solution:
1) The equation is balanced as written.
2) Determine moles of each reactant:
C4H9OH ---> 15.0 g / 74.122 gmol = 0.20237 mol
NaBr ---> 22.4 g / 102.894 gmol = 0.2177 mol
H2SO4 ---> 32.7 g / 98.0768 g/mol = 0.3334 mol
3) Dividing the mole amount of each substance by the respective coefficient of the balanced equation will identify the limiting reagent since it will have the lowest value answer. All the coefficients have a value of 1, therefore it is seen that C4H9OH is the limiting reagent.
4) Determine theoretical yield of C4H9Br:
Since I have already calculated the moles of C4H9OH, that is the value I start with.
0.20237 mol C4H9OH 1 mol C4H9Br 137.0191 g C4H9Br –––––––––––––––––– x ––––––––––––– x ––––––––––––––– = 27.7286 g C4H9Br 1 mol C4H9OH 1 mol C4H9Br
5) Determine percent yield:
(17.1 g / 27.7286 g) * 100 = 61.7%
Problem #22: Consder the following reaction:
Fe + O2 ---> Fe2O3
Given that 50.0 g of Fe is present initially, determine the limiting reagent and the excess reagent.
Solution:
The are no calculations associated with the solution to this problem.Since the amount of oxygen is not stated, the convention used in chemistry is to assume it is present in excess.
This makes the iron the limiting reagent.
Problem #23: Determine the limiting and excess reagents for the following reaction:
Mg(OH)2 + 2HCl ---> MgCl2 + H2O
And given the following amounts: 0.0218 mole of Mg(OH)2 and 0.0436 mole of HCl
Solution:
1) Assume Mg(OH)2 is limiting. How much MgCl2 is produced?
0.0218 mol Mg(OH)2 1 mol MgCl2 ––––––––––––––––– x ––––––––––– = 0.0218 mol MgCl2 1 mol Mg(OH)2
2) Assume HCl is limiting. How much MgCl2 is produced?
0.0436 mol HCl 1 mol MgCl2 ––––––––––––– x –––––––––– = 0.0218 mol MgCl2 2 mol HCl
3) The two reactants run out simultaneously. In this case, some teachers (& textbooks) consider there to be two limiting reagents. Other teachers (& textbooks) tend to say there is not a limiting reagent. Both have a point, so the ChemTeam (when he was in the classroom) would point out this situation, say be aware of it in the future and then tell his classes there would not be a question about this point on the test.
Problem #24: Someday.
Problem #25: Someday.
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