Unit 7 - Stoichiometry Flashcards Preview

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Flashcards in Unit 7 - Stoichiometry Deck (12):

Excess reagent/reactant

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


Limiting Reagant/Reactant

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.)


Theoretical Yeild

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


Actual Yeild

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


Percent Yeild

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


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

3.75 moles O₂ x (2 moles Al₂O₃/ 3 moles O₂) =

2.50 moles Al₂O₃


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

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

21 moles CO₂


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

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


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

3.75 moles Cl₂ x ( 2moles NaCl/1 mole Cl₂ ) x ( 58.5g NaCl / 1 mol NaCl =


438.75 g NaCl


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

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

4.87 moles Na.


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?

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.


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.

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 %