Stoichiometry and Stoichiometric Calculations

Stoichiometry is the quantitative relationships between the reactants and products in a chemical reaction.

It involves the calculation of the amounts (masses, volumes, moles etc) of reactants and products involved in a chemical reaction.

Stoichiometric Calculations

For all volume calculations, use the fact that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters or click here to learn more.

Mole-Mole Calculations:

In the reaction 2H₂ + O₂ → 2H₂O, how many moles of water will be produced from 3 moles of hydrogen gas?

Solution

i. Chemical equation:
2H₂ + O₂ → 2H₂O
From the equation:
2 moles of hydrogen gas produces 2 moles of water
3 moles of hydrogen gas produces ? moles of water.

Moles of water = 3mol x 2mol
                     2mol
		= 3mol 
Therefore, 3 moles of water will be produced.

The reaction 2Al + 3Cl₂ → 2AlCl₃ is given. How many moles of chlorine are needed to react with 4 moles of aluminum?

Solution

2Al + 3Cl₂ → 2AlCl₃

2 mol of Al → 3mol of Cl₂ 
4 mol of Al → ? mol of Cl₂

Moles of chlorine = 4mol x 3mol
                       2mol
		  = 6mol
6 moles of chlorine are needed

For the reaction N₂ + 3H₂ → 2NH₃, how many moles of ammonia will be produced from 1.5 moles of nitrogen?

Solution

N₂ + 3H₂ → 2NH₃
From the equation:
1 mole of N₂ → 2 moles of NH₃
1.5 moles of N₂ → ? moles of NH₃
Moles of NH₃ = 1.5mol x 2mol
                 1mol
	     = 3mol 
3 moles of ammonia will be produced.

In the first stage of rusting, iron reacts with oxygen according to the equation, 4Fe + 3O₂ → 2Fe₂O₃, determine the moles of iron (II) oxide produced from 6 moles of iron.

Solution

The equation:  4Fe + 3O₂ → 2Fe₂O₃
 From the equation:
  4 mol of Fe → 2 mol of 2Fe₂O₃
  6mol of Fe → ? mol of 2Fe₂O₃
  Moles of 2Fe₂O₃ = 6mol x 2mol
                      4mol
		  = 3mol 
Therefore, 3 moles of 2Fe₂O₃ will be produced.

In the reaction C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O, how many moles of oxygen are required to react with 2 moles of ethanol?

Solution

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
From the equation:
1 mole of C₂H₅OH → 3moles of O₂
2 moles of C₂H₅OH → ? moles of O₂

Moles of O₂ = 2mol x 3mol
                1mol
            = 6mol
Therefore, 6 moles are required.

When Kaluma heated strongly unknown substance W, colorless gas evolved which turned lime water milky and the remaining substance was white in color. After contacting his teacher, the teacher said the unknown substance was calcium carbonate.
(a) Write the balanced chemical equation for thermal decomposition reaction took place.
(b) How many moles of calcium oxide will be produced from 5.56 moles of calcium carbonate?

Solution
```
(a) CaCO₃ → CaO + CO₂
(b) From the equation:
1 mole of CaCO₃ → 1 mole of CaO 
5.56 moles of CaCO₃ → ? mole of CaO

Mole of CaO = 5.56mol x 1mol
                   1mol 
				   
	    = 5.56mol
Therefore, 5.56 moles will be produced.
```

Mole-Mass Calculations:

How many grams of sodium chloride (NaCl) will be produced from 2 moles of sodium (Na) reacting with excess chlorine (Cl₂)?

Solution

(i) Chemical equation:
2Na + Cl₂ → 2NaCl
2 mol Na → 2 mol NaCl 
2 mol Na → ? mol of NaCl

Moles of NaCl = 2mol x 2mol
                    2mol
	          = 2mol 
(ii) Molar mass of NaCl = 23 + 35.5 = 58.5g/mol			  
From number of moles = mass of the substance
	                     molar mass
Mass of the substance = number of moles x molar Mass 
                      = 2mol x 58.5g/mol 
                      = 117g

Given the reaction 2H₂ + O₂ → 2H₂O, how many grams of water can be produced from 5 moles of hydrogen gas?

Solution

From the equation:
2 moles of H₂ → 2moles of H₂O
5 moles of H₂ → ? moles of H₂O

Number of moles H₂O = 5 mol x 2 mol
                       2mol
	         = 5mol 
Molar mass of H₂ = (1 x 2) + 16 = 18g/mol 
Mass of H₂O = 5mol x 18g/mol
           = 90g

For the reaction 2K + Cl₂ → 2KCl, how much potassium chloride in grams can be produced from 3 moles of potassium?

Solution

Molar mass KCl = 39 + 35.5 = 74.5g/mol 
Mass of KCl = 3 mol x 74.5g/mol
            = 223.5g

Based on Fe₂O₃ + 3CO → 2Fe + 3CO₂, how many grams of iron are produced from 2 moles of iron (III) oxide?

Solution

From the equation:
1 mole of Fe₂O₃ →  2 moles of Fe
? mol of Fe₂O₃ → 2 moles of Fe

Number of moles of Fe₂O₃ = 1mol x 2mol
                             2mol
		         = 1mol 
	Molar mass of Fe₂O₃ = (56 x 2) + (16 x 3)
	                   = 160g/mol 
	Mass of Fe₂O₃ = 1mol x 160g/mol 
                      = 160g

Mass-Mass Calculations:

In the reaction C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, how many grams of water are produced from 88 grams of propane?

Solution

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Molar mass of C₃H₈ = (12 x 3) + (1 x 8)
                     = 44g/mol
The mass of C₃H₈ = 1 mol x 44g/mol
                 = 44g
				 
Molar mass of H₂O = 18g/mol
Mass of H₂O = 4mol x 18gmol
            = 72g
			
From the equation: 
44g of C₃H₈   → 72g of H₂  
88g of C₃H₈   →  ? g of H₂
 
  Mass of water =  88g x 72g 
	              44g
		= 144g

For 2Fe₂O₃ + 3C → 4Fe + 3CO₂, how many grams of iron can be produced from 160 grams of iron (III) oxide?

Solution

Mass of iron (Fe) = 4mol x 56g/mol
                  = 224g 			 
Mass of iron(III)oxide (Fe₂O₃) = 2mol x 160g/mol 
                               = 320g
Therefore:
320g of Fe₂O₃ → 224g of Fe 
160g of Fe₂O₃ → ?g of Fe

Mass of Fe produced = 160g x 224g
                        320g	
                     = 112g

Given NH₃ + HCl → NH₄Cl, calculate the grams of ammonium chloride produced from 34 grams of ammonia.

Solution

Molar mass of NH₃ = 14 + 3
                  = 17g/mol
Mass from the equation = 1mol x 17g/mol 
                       = 17g

Molar mass of NH₄Cl = 14 + (1 x 4) + 35.5
                    = 53.5g/mol
Mass of NH₄Cl from the eqn. = 1 mol x 53.4g/mol
                            = 53.5g

  17g of NH₃ →  53.5 of NH₄Cl 
  34g of NH₃ →  ?g of NH₄Cl 
  
  Mass of NH₄Cl  =  34g x 53.5g
                       17g
		= 107g

Mole-Volume Calculations:

For the reaction CaCO₃ + 2HCl → CaCl₂ + H₂ O + CO₂, how many liters of carbon dioxide at STP are produced from 2 moles of calcium carbonate?

Solution

CaCO₃ + 2HCl → CaCl₂ + H₂ O + CO₂
1 mole of CaCO₃ → 1 mole of CO₂ = 22.4dm³/mol at STP
2 moles of CaCO₃ → ? dm³ of CO₂
Volume of CO₂ = 2mol x 22.4dm³/mol
                   1mol
	     =  44.8dm³
Therefore, 44.8litres of carbon dioxide will be produced.

In the reaction C₂H₄ + 3O₂ → 2CO₂ + 2H₂O, how many liters of oxygen at STP are required to react with 4.25 mole of ethylene?

Solution

From the equation:
 1 mol of C₂H₄ →   67.2dm³ of oxygen     ←( 3 mol x 22.4dm³/mol )
4.25 mol of C₂H₄ → ? dm³ of oxygen

Volume of oxygen = 4.25mol x 67.2dm³
                 = 285.6dm³
	        = 285.6litres

Mass-Volume Calculations:

In the complete combustion of 112 grams of butane (2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂ O), how many liters of water vapour at STP are produced?

Solution

Molar mass of butane = (12 x 4) + (10 x 1)
                     = 58g/mol 
From the equation:
 Mass of butane = number of moles x molar mass
                = 2mol x 58g/mol 
                = 116g
 Volume of water vapour = 10mol x 22.4dm³
                        = 224dm³

Therefore:					   
116g of butane → 224dm³ of water vapour,
112g of butane →  ? dm³ of water vapour 

Volume of water vapour = 112g x 224dm³
                             116g	
                       = 216.28dm³	
216.28 litres of water vapour were produced.

Given CaCO₃ + 2HCl → CaCl₂ + H₂ O + CO₂, how many liters of carbon dioxide at STP are produced from 100 grams of calcium carbonate?

Volume-Volume Calculations:

In 2NH₃ + 3CuO → 3Cu + N₂+ 3H₂O, if 11.2 liters of ammonia at STP react with excess copper(II) oxide, how many liters of nitrogen at STP are produced?

Solution

2NH₃ + 3CuO → 3Cu + N₂+ 3H₂O
Number of moles = volume of the gas
                     molar volume
					 
At STP molar volume = 22.4dm³/mol

Therefore:
Volume of a gas = number of moles x molar Volume
From the equation:
 Volume of ammonia (NH₃)  = 2mol x 22.4dm³/mol 
                          = 44.8dm³
Volume of nitrogen (N₂) = 1mol x 22.4dm³
                         = 22.4dm³

22.4dm³ of N₂ → 44.8dm³ of NH₃
11.2dm³ of N₂ → ? dm³ of NH₃
Volume of nitrogen = 11.2dm³ x 44.8dm³
                           22.4dm³
The of nitrogen is 22.4 litres.

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