# STOICHIOMETRIC CALCULATIONS

STOICHIOMETRIC CALCULATIONS

Amount of substance used or produced in a chemical reaction.

Calculations to determine the amount of reactants or products of a chemical reaction.

Example:

A + B ^{___________}> C ,

A and B are reactants, C is product, therefore, 1mol A reacted with 1mol B to give 1mol C

Mole ratio can be written as mol **A / mol B = 1/1 or mol A/mol C = 1/1 or mol B / mol C = 1/1**

Therefore, this means ; 1mole B = 1mole A = 1mole C

If A is present in more moles than B then A is excess reagent or reactant, and B will be the LIMITING reactant.

This means moles of B determine amount of C product expected from reaction.

% yield of product C = AMOUNT of C actually obtained / [amount of C expected] x 100

Example :

A + 3B^{ __________}> 2C

Mole Ratios ;

**A / mol B = 1/3 or mol A/mol C = 1/2 or mol B / mol C = 3/2**

Example

- 2.0g of tin was burned up in oxygen gas to form Tin [II]Oxide. Calculate the % yield if 2.0g of product was obtained.

ANSWER:

It is obvious oxygen gas is excess reagent and tin is limiting reagent.

2Sn + O_{2} ^{___}> **2**SnO

Molar mass of Sn = 118.71g/mol

Moles of Sn = (1.0mol/118.71g) x 2.0g = 0.01685 mol

From reaction equation,

2 mol Sn should give 2 mol SnO

Therefore 0.01685 mol Sn should give ,

(0.01685mol Sn / 2mol Sn ) x 2mol SnO = (**2**/2) x 0.01685 mol SnO = **0.01685 mol SnO**

**Molar mass SnO = [16 + 118.71] g/mol = 134.71 g/mol**

**Mass of SnO expected = 0.01685mol x 134.71g/1mol = 2.27g**

**% yield = [amount obtained / expected among] x 100**

** = [ 2.0/2.27 ] x 100 **

** = 88.11%**

**2. A reaction between 4 g Nitrogen gas and 4 g oxygen gas gave 60g of N _{2}O_{4 }. Calculate the percent yield.**

** **

**ANSWER**

**N _{2} + 2O_{2 } ^{______}> N_{2}O_{4}**

**Moles of nitrogen used = [1mol/(2 x 14g)] x 4.0g = 28.0mol**

**Moles of oxygen gas used = [1 mol / (2 x 16 g)] x 4g = 32.0mol**

**From the reaction equation 1 mol N**_{2}** should give 1 mol N**_{2}**O**_{4}** , **

**therefore 28.0mol N**_{2}** should give 28.0 mol N**_{2}**O**_{4}

**From the reaction equation 2 mol O _{2} should give 1 mol N_{2}O_{4} , **

**Therefore 32 mol O**_{2}** should give;**

** ( 32 mol O**_{2}** /2 mol O****2 **** ) x 1 mol N**_{2}**O**_{4}** = (½) 32 mol N**_{2}**O**_{4}** = 16mol **

**Oxygen gives smaller moles of product N _{2}O_{4} = LIMITING REAGENT, Therefore N_{2} is EXCESS REAGENT.**

**We use limiting reagent to calculate expected amount of product which is 16mol N _{2}O_{4}. **

**Molar mass N _{2}O_{4} = 2 x 14 + 4 x 16 = 28 + 64 = 92 g/mol**

**Moles of 60g N _{2}O_{4} actually obtained = (1mol/92g) x 60 g = 0.6522 mol **

**% yield = [mole product actually obtained(0.6522mol) / mole product expected(16mol)] x 100**

** = (0.6522/16) x 100 **

** = 4 %**

**Calculate the moles of the excess reagent that did not participate in the chemical reaction.**

**The reaction LIMIT is 16mol of product. Theoretical excess would be [28-16] mol = 12mol N _{2}O_{4}**

**Reaction equation shows moles of N _{2} = moles of N_{2}O_{4}, therefore 12 mole N_{2}O_{4} is equivalent to 12 mol N_{2}**

**Therefore 12 mol of N _{2} did not react.**

_{ }**3. How much Potassium Chlorate decomposes to oxygen gas 100L of oxygen gas at STP.**

**2****KClO _{3} ^{__________}> 2KCl + 3O_{2}**

ANSWER:

**At STP is 22.41 L contains 1 mol O _{2 }, therefore 100L will be (1mol/22.41 L) x 100 L = 4.46mol**

**From the above equation, 2 mol KClO _{3} should give 3 mole O_{2} **

**Therefore,**** 4.46mol O _{2} equivalent to**

** (4.46mol O _{2} / 3mol O_{2} ) x 2moles KClO_{3} = [2/3] x 4.46 mol = 2.97 mol KClO_{3}**

**Mass = 2.97mol x molar mass KClO _{3 }(122.55g/mol) = 363.97g**

PS. You can also use the mole ratio set up equations to solve these problems above and get the same answers.