Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1. 2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube? 1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2. Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3. Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen 2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
Stoichiometry is the quantitative study of the reactants and products involved in a chemical reaction. Stoichiometry is also the calculation of products and reactants in a chemical reaction. It deals with the number of products and reactants involved in a chemical reaction.
Importance of Chemical Stoichiometry
It helps chemists to use balanced chemical equations to determine the moles of reactants and products in a chemical reaction.
It helps to determine the mass of reactants and products required or present in a chemical reaction.
It helps chemists measure the molecular weight of substances.
It helps to measure the formulae.
Basic Concepts of Stoichiometric Reaction
In a stoichiometric reaction, there are some basic things to note to make it easier to understand. We can get the pictures of these things by considering the equation below:
3Fe(S) + 4H2O(I) → Fe3O4(S) + 4H2(g)
The stoichiometric coefficient of this chemical reaction represents the coefficient of each element or compound involved in the chemical reaction.
So, 3 moles of Fe (iron) react with 4 moles of water to produce one mole of iron(II)diiron(III)oxide and 4 moles of hydrogen gas.
The chemical equation can also tell us how many grams of the reactants and products are involved. So we can say that 168g of iron reacts with 72g of water to give 231g of iron(II)diiron(III)oxide and 8g of hydrogen gas.
If the reactants and products are in a gaseous state, then we can consider the molar volume of gases which is 22.4dm3 (22.4L) for one mole of any gas.
N2(g) + 3H2(g) → 2NH3(g)
22.4dm3 of nitrogen reacts with 672dm3 (3x22.4) of hydrogen to give 44.8dm3 (2x22.4) of ammonia. From the two chemical equations, we were able to interpret the chemical reaction using their stoichiometric coefficients.
Stoichiometry Questions and Answers
Question one:
Calculate the mass of potassium hydroxide required to prepare 500cm3 of 0.2mole per dm3 (per a litre) of solution. (0.2mole per dm3 = 0.2M) (K=39, O=16, H=1)
Solution
Molar mass of KOH = 56g/mol
Volume required = 500cm3
Volume given = 1000cm3
Molarity (concentration in mole per dm3) = 0.2M
Concentration in mass per dm3 of the given solution = 0.2x56 = 11.2g/dm3
Since 1000cm3 of KOH solution contains = 11.2g
500cm3 of KOH solution will contain
= 11.2/1000 x 500 = 5.6g
Question two:
What is the volume of 10M H2SO4 that will be required to prepare 4M in 500cm3 of H2SO4?
Solution
In this case, we are going to use the dilution formula
C1V1= C2V2
Where
C1 = 10M
V1 = ??
C2 = 4M
V2 = 500cm3
10 x V1 = 4x500
V1 = 2000/10 = 200cm3
Question three:
5g of a solid mixture containing NaCl is dissolved in water and treated with an excess of AgNO3, resulting in the precipitation of 2.1g of AgCl. What percentage of the mixture was AgCl?
(Na =23, N=14, O =16, Cl =35.5, Ag =108)
NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)
Solution
From the chemical equation,
143.5g of AgCl required = 58.5g of NaCl
2.1g of AgCl will require = 58.5/143.5 x 2.1
= 0.86g of NaCl
So 0.86g of NaCl reacted with AgNO3 since the mass of the solid mixture is 5g
Then
Percentage of the mixture that was AgCl = 0.86/5 x 100
17.1%
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