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

1.      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 + O₂ ^___> 2SnO

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 N₂O₄ . Calculate the percent yield.

 

ANSWER

N₂ + 2O₂    ^______>   N₂O₄

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 N₂ should give 1 mol N₂O₄ ,

therefore 28.0mol N₂ should give 28.0 mol N₂O₄

From the reaction equation 2 mol O₂ should give 1 mol N₂O₄ , 

Therefore 32 mol O₂ should give;

(  32 mol O₂ /2 mol O2     )  x 1 mol N₂O₄   =  (½) 32 mol N₂O₄ = 16mol

Oxygen gives smaller moles of product N₂O₄ = LIMITING REAGENT, Therefore N₂ is EXCESS REAGENT.

We use limiting reagent to calculate expected amount of product which is 16mol N₂O₄.

Molar mass N₂O₄ = 2 x 14 + 4 x 16 = 28 + 64 = 92 g/mol

Moles of 60g N₂O₄ 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 N₂O₄

Reaction equation shows moles of N₂ = moles of N₂O₄, therefore 12 mole N₂O₄ is equivalent to 12 mol N₂

Therefore 12 mol of N₂ did not react.

   3.      How much Potassium Chlorate decomposes to oxygen gas 100L of oxygen gas at STP.

2KClO₃ ^__________> 2KCl   +   3O₂

ANSWER:

At STP is 22.41 L contains 1 mol O₂     , therefore 100L will be (1mol/22.41 L) x 100 L  =  4.46mol

From the above equation,    2 mol KClO₃ should give 3 mole O₂

Therefore, 4.46mol O₂ equivalent to

(4.46mol O₂ / 3mol O₂ ) x  2moles KClO₃ = [2/3] x 4.46 mol = 2.97 mol KClO₃

Mass = 2.97mol x molar mass KClO₃   (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.