Limiting Reactants: Quiz

Limiting Reactants: Quiz

7 questions

1. Question 1EasyWhat is the limiting reactant in a reaction?

   AThe reactant with the smallest massBThe reactant that is used up firstCThe reactant with the smallest coefficientDThe reactant that is left over at the end
   Show answer

   Answer: The reactant that is used up first

   The limiting reactant runs out first, so it decides how much product can form. The reactant left over at the end is the excess reactant.

2. Question 2MediumIn $\ce{2H2 + O2 -> 2H2O}$, 1.0 mol of $\ce{H2}$ is mixed with 0.6 mol of $\ce{O2}$. Which is the limiting reactant?

   A$\ce{H2}$B$\ce{O2}$CNeither; they react exactlyDBoth run out at the same time
   Show answer

   Answer: $\ce{H2}$

   1.0 mol of $\ce{H2}$ needs $1.0 \times \dfrac{1}{2} = 0.50$ mol of $\ce{O2}$. There is 0.6 mol available, so oxygen is in excess and hydrogen runs out first, even though there are fewer moles of oxygen.

3. Question 3MediumIn $\ce{N2 + 3H2 -> 2NH3}$, 2.0 mol of $\ce{N2}$ is mixed with 3.0 mol of $\ce{H2}$. How many moles of $\ce{NH3}$ can form?

   A2.0 molB3.0 molC4.0 molD6.0 mol
   Show answer

   Answer: 2.0 mol

   All the $\ce{N2}$ would need 6.0 mol of $\ce{H2}$, but only 3.0 mol is available, so $\ce{H2}$ is limiting: $3.0 \times \dfrac{2}{3} = 2.0$ mol of $\ce{NH3}$. The answer 4.0 mol uses the excess reactant.

4. Question 4EasyThe theoretical yield of a reaction is 50.0 g, and 42.5 g of product is collected. What is the percent yield?

   A7.50%B85.0%C92.5%D118%
   Show answer

   Answer: 85.0%

   Percent yield = $\dfrac{42.5}{50.0} \times 100\% = 85.0\%$. The answer 118% divides the wrong way round.

5. Question 5MediumIn $\ce{2H2 + O2 -> 2H2O}$, 4.0 mol of $\ce{H2}$ reacts with 3.0 mol of $\ce{O2}$. How much of the excess reactant is left over?

   A1.0 mol of $\ce{O2}$B2.0 mol of $\ce{O2}$C1.0 mol of $\ce{H2}$DNothing is left over
   Show answer

   Answer: 1.0 mol of $\ce{O2}$

   4.0 mol of $\ce{H2}$ needs only 2.0 mol of $\ce{O2}$, so hydrogen is limiting. Oxygen left over = 3.0 − 2.0 = 1.0 mol.

6. Question 6Hard16.0 g of methane burns with 48.0 g of oxygen: $\ce{CH4 + 2O2 -> CO2 + 2H2O}$. What mass of $\ce{CO2}$ can form? (C = 12.01, H = 1.008, O = 16.00)

   A33.0 gB43.9 gC64.0 gD66.0 g
   Show answer

   Answer: 33.0 g

   Moles: $\ce{CH4}$ = 16.0 ÷ 16.04 = 0.9974 mol; $\ce{O2}$ = 48.0 ÷ 32.00 = 1.50 mol. The methane would need 1.995 mol of $\ce{O2}$, so oxygen is limiting. $\ce{CO2}$ = 1.50 ÷ 2 = 0.750 mol, and 0.750 × 44.01 = 33.0 g. The answer 43.9 g uses the excess reactant (methane).

7. Question 7EasyA student reports a percent yield of 112%. What is the most likely explanation?

   AThe reaction produced extra product from nothingBThe product was still wet or contained impuritiesCThe limiting reactant was used in excessDPercent yields are always above 100%
   Show answer

   Answer: The product was still wet or contained impurities

   Mass cannot be created, so a yield above 100% signals an error, most often a product that still contains water (solvent) or other substances.