Honors - Unit 6 - Stoichiometry

Question 1

Excess reagent/reactant

Answer 1

The reactant/reagant which is not completely used up during a reaction.

Question 2

Limiting Reagant/Reactant

Answer 2

Determines the amount of product that can be formed.

The reactant/reagant that is completely used up during a reaction.

(In the attached picture, beef patties are limiting how many hamburgers we can make.)

Question 3

Theoretical Yeild

Answer 3

The maximum amount of product that is CALCULATED to be formed from the given amounts of reactants.

Question 4

Actual Yeild

Answer 4

The amount of product which is actually formed when the reaction is carried out in the laboratory.

Question 5

Percent Yeild

Answer 5

A measure of the efficiency of a reaction carried out in the laboratory.

Question 6

In the equation:
4Al + 3O₂→ 2Al₂O₃,
how many moles of Al₂O₃will be produced if there are 3.75 moles of O₂?

Answer 6

3. 75 moles O₂ x (2 moles Al₂O₃/ 3 moles O₂) =
4. 50 moles Al₂O₃

Question 7

In the equation:
C₃H₈ + 5O₂→ 3CO₂+ 4H₂O
How many moles of CO₂are produced when 7 moles of C₃H₈ are burned?

Answer 7

7 moles C₃H₈ x (3 moles CO₂ / 1 mole C₃H₈) =

21 moles CO₂

Question 8

In the equation:
NH₄NO₃ → N₂O + 2H₂O
How many grams of H₂O are produced if you are given 50 g of NH₄NO₃?

Answer 8

50 g NH₄NO₃ x (1 mole NH₄NO₃ /80.04 gNH₄NO₃) x
(2 moles H₂O / 1 mole NH₄NO₃) x (18.02 g H₂O/ 1 mol H₂O) =

22.5 g H₂O

Question 9

In the equation:
2Na + Cl₂→ 2NaCl
How many grams of NaCl are produced from 3.75 moles of Cl₂?

Answer 9

3. 75 moles Cl₂ x ( 2moles NaCl/1 mole Cl₂ ) x ( 58.5g NaCl / 1 mol NaCl =
4. 75 g NaCl

Question 10

In the equation:
2Na + Cl₂→ 2NaCl
How many moles of Na are needed if 285 g of NaCl are produced?

Answer 10

285g NaCl x (1 mole NaCl/ 58.5 g NaCl) x (2 moles Na / 2 moles NaCl) =

4.87 moles Na.

Question 11

Given the following reaction: S₈ + 4Cl₂ → 4S₂Cl₂. If there is 200g of sulfur and 100g of chlorine, what mass of disulfur dichloride will be produced?

Answer 11

1. Perform a mass-to-mass calculation between sulfur and disulfur dichloride.

200g S₈ x (1mol S₈ / 256.5g S₈) x (4mol S₂Cl₂ / 1 mol S₈) x (135g S₂Cl₂ / 1 mol S₂Cl₂) = 421g S₂Cl₂.

2. Perform a mass-to-mass calculation between chlorine and disulfur dichloride.

100g Cl₂ x (1mol Cl₂ / 70.91g Cl₂) x (4mol S₂Cl₂ / 4 mol Cl₂) x (135g S₂Cl₂ / 1 mol S₂Cl₂) = 190.4g S₂Cl₂.

The limiting reactant is chlorine (Cl₂ ) because… it produced only 190.4g S₂Cl₂.

The excess reactant is sulfur (S₈) because… it would have produced 421g S₂Cl₂.

Question 12

Calculate the % Yield of solid silver chromate produced in the following reaction:
K₂CrO₄ + 2AgNO₃ → Ag₂CrO₄ + 2KNO₃
In the reaction there was .500g of the limiting reactant AgNO₃. In the actual experiment, .455g of Ag₂CrO₄ was produced.

Answer 12

Calculating the Theoretical Yield of Ag₂CrO₄ that was produced:

0.500g AgNO₃ x (1 mole AgNO₃ / 169.9g AgNO₃) x
(1mole Ag₂CrO₄ / 2 mole AgNO₃) x (331.7 g Ag₂CrO₄ / 1 mol Ag₂CrO₄) =0 .488g Ag₂CrO₄.

Find the ratio of the actual yield (.455g) to the theoretical yield (.488g) …

% Yield = (.455g Ag₂CrO₄ ÷ .488g Ag₂CrO₄ ) x 100 =
93.2 %