1.3 Reacting masses and volumes

1.3 Reacting masses and volumes

Solve the following Practice Questions on Reacting masses using balanced Chemical equations

Q.1 20 g Magnesium ribbon is burnt in air with white light. Calculate the mass of Magnesium oxide (MgO) formed in this reaction.

Hint: Write the balanced chemical equation for the reaction between Mg and Oxygen gas.

2Mg(s) + O2 (g) → 2MgO(s)

Use the Molar mass of Mg and MgO according to balanced equation and calculate the mass of MgO.

Mr of Mg= 24 g/mol

Mr of O = 16 g/ mol

So MgO = 24+16 = 40 g/mol

2Mg(24×2) + O2 (g) → 2MgO(40×2)

So According to balanced equation,

48 g Mg forms MgO = 80 g

Hence 20 g Mg will form MgO = (20X80)/48

=33.33 g thus 20 g Mg will form 33.33 g MgO

We did not take Oxygen in to account because the Oxygen is not a limiting reactant and it is enough in the reaction.

Q2. 25 g Li2O(s) is reacted with 100 cm3 of  0.75mole/dm HCl solution as per below equation:

Li2O+ 2HCl(aq) → 2LiCl (aq) + H2O(l)

Calculate the amount of LiCl formed in the reaction.

Q3. 2.5 moles of MgO  reacts with  2.5 moles of HCl. Determine the limiting reactant and calculate the amount of  MgCl2 formed in the reaction.

MgO(s) + 2HCl (aq) → MgCl2(aq) + H2O(l)

Q.4  150g Phosphorus reacts with 200 g Oxygen gas , determine the limiting reactant and calculate the mass of Phosphorus oxide(P4O10 ) formed in this reaction.

Limiting Reactant

The reagent which is finished earlier than other reactants in a reaction is called limiting reagent.

As soon as the limiting reagent finishes, the reaction stops and no more product forms.

Only one reactant can be limiting in a chemical reaction.

Always calculate the amount of product by using limiting reactant and not the excess reactant because to form the product you need all the reactants.

For example: 20 g Magnesium burns in 10 g of Oxygen to form Magnesium oxide. Determine the limiting reactant and calculate the mass of magnesium oxide formed in the reaction.

Mg ribbon burning

Firstly write the chemical equation for the above reaction.

Mg + O2  –> MgO

Balance the equation

2Mg +O2 –>2MgO

Ar of Mg is 24 g/mol

Ar of O is 16 g/mol

Mr of MgO is 24+16=40 g/mol

48 g Mg needs Oxygen= 32 g

20 g Mg Needs Oxygen= x

x=( 20*32)/48

x= 13.3 g

so 13.3 g Oxygen is needed in the reaction but we only have 10 g oxygen in the reaction thus Oxygen is limiting reactant.

Now we need to calculate the mass of magnesium oxide formed so again we will use the balanced equation and only compare the limiting reactant to calculate the product as informed earlier.

32 g Oxygen forms MgO = 80 g

10 g Oxygen forms MgO = x = 80*10/32


Definition of Avogadro’s Law : 

“The equal volumes of all gases have equal number of moles” under standard conditions of temperature and pressure.

Photo of Avogadro

Standard temperature is 273 K or 00 Celsius

Standard Pressure is 100 kPa

Thus 1 mole of each gas at 273 K temperature and 100 kPa pressure,  will have 22.7 dm3.

22.7 dm3 is called molar gas volume.

Numerically Volume V is directly proportional to number of moles.

V ∝ n

V=kn where k is proportionality constant.


1 mole of O2 and 1 mole of CO2 will have 22.7 dm3 volume at 273 K temperature and 100 kPa pressure. 6.02X1023

Photo of Avogadro

Ideal Gas: 

The hypothetical gas which obeys gas laws as per the equation PV= nRT where 

V=  Volume, P= Pressure, T= Kelvin Temperature, and n=the number of moles of gas present.

Real gases show deviation from Ideal gas equation because Real gases do have some volume and gas particles also attract each other if compressed.


It is the number of moles of a solute in 1 dm-3 of a solution. It’s units are Moldm-3.

Acid-Base titration: 

Titration is an experimental technique to find an unknown concentration of acid or alkali if the other’s concentration is already known.

The standard solution is the one whose concentration we know..

When the neutralization occurs, we call it Equivalence point.

An Indicator is required to know the equivalence point or the completion of the reaction.


The acid is filled in the burette while the base is filled in conical flask using a pipette. Few drops of an indicator are mixed in the conical flask. The acid should add drop wise drop from burette to the conical flask till the equivalence point which is clear by the colour change.

I hope by now you have learned the subtopic 1.3 reacting masses and volumes.

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