### Stoichiometry Worksheet, Balanced Equations, Grams, Reactant, Limiting Reactant

The key terms in this Chemistry course include Stoichiometry Worksheet, Balanced Equations, Grams, Reactant, Limiting Reactant, Excess Reactant, Product, Reaction, Theoretical yield, Calculations, P4, Cl2, PCl_{3}, H3PO4, Fe2O3,H2O, CO, Fe, CO2, P4O10, Stoichiometry Worksheet – Chemistry.

**Using the three balanced equations below, do 9.a through 9.d for each:**- How many grams of the first product can be produced from 20.0 grams of the first reactant and 20.0 grams of the second reactant?
- Which reactant is the limiting reactant?
- How many grams of the excess reactant are reacted?
- How many grams of the excess reactant are left unreacted?

__P4 + 6Cl____2__ **➡** __4PCl3__** **__P4O10 + 6H2O__ **➡**** **__4H3PO4__** **__Fe2O3 + 3CO __ **➡** __2Fe+ 3CO2__

9 a. | 9 a. | 9 a. | ||

9 b. 9 c. | 9 b. 9 c. | 9 b. 9 c. | ||

9 d. | 9 d. | 9 d. |

P4 + 6Cl2 ** ➡ ** 4PCl3

How many grams of the first product can be produced from 20.0 grams of the first reactant and 20.0 grams of the second reactant?

68.71 g 2^{nd} reactant needed.

**A student carried out the reaction shown by reacting 10.0g P**_{4 }with excess Cl_{2 }and obtained 35.0g of PCl_{3}.

**P _{4 }+ 6Cl_{2} ➡ 4PCl_{3}**

What is the theoretical yield of PCl_{3}?

Calculate the % yield ofPCl_{3}

**Calculations**

**10. A student carried out the reaction shown by reacting 10.0g P _{4 }with excess Cl_{2 }and obtained 35.0g of PCl_{3}.**

P_{4} + 6Cl_{2} **➡** 4PCl_{3}

Theoretical Yield of PCl_{3}

10 x (1/123.9) x (4/1) x (137.88/1) = 44.51g

% Yield of PCl_{3}

(35/44.51) x 100 = 78.63 %

**Using the three balanced equations below, do 9.a through 9.d for each:**- How many grams of the first product can be produced from 20.0 grams of the first reactant and 20.0 grams of the second reactant?
- Which reactant is the limiting reactant?
- How many grams of the excess reactant are reacted?
- How many grams of the excess reactant are left unreacted?

** P4 + 6Cl**2

**➡**4PCl3P4O10 + 6H2O

**➡**4H3PO4Fe2O3 + 3CO

**➡**

__2Fe+ 3CO2__9 a. | 9 a. | 9 a. | ||

9 b. 9 c. | 9 b. 9 c. | 9 b. 9 c. | ||

9 d. | 9 d. | 9 d. |

P4 + 6Cl2 **➡ ** 4PCl3

68.71 g 2^{nd} reactant needed.

**A student carried out the reaction shown by reacting 10.0g P**_{4 }with excess Cl_{2 }and obtained 35.0g of PCl_{3}.

**P _{4 }+ 6Cl_{2} ➡ 4PCl_{3}**

What is the theoretical yield of PCl_{3}?

Calculate the % yield ofPCl_{3}

**Calculations**

**10. A student carried out the reaction shown by reacting 10.0g P _{4 }with excess Cl_{2 }and obtained 35.0g of PCl_{3}.**

P_{4} + 6Cl_{2} **➡** 4PCl_{3}

Theoretical Yield of PCl_{3}

10 x (1/123.9) x (4/1) x (137.88/1) = 44.51g

% Yield of PCl_{3}

(35/44.51) x 100 = 78.63 %