Molar ratio question are very common in BMAT exams. This involves the use of Moles and Reaction equations. 

It is best to use an example to understand molar ratios. 


Example 1:
Sodium reacts with water to form sodium hydroxide and water. 4.76 grams of sodium hydroxide was formed. Find the mass of sodium used. 

2Na + 2H₂O  →  2NaOH + H₂
(Ar of Na = 23, O = 16, H = 1)

We first need to find the Mr of NaOH. 
(23 x 1) + (16 x 1) + (1 x 1) = 23 + 16 + 1 = 40

n = M ÷ Mr
n = 4.76 ÷ 40 = 0.119

The reaction equation above shows that 2 moles of Na forms 2 moles of NaOH (highlighted in red colour).

Molar ratio = 2 : 2 = 1 : 1
Therefore, moles of Na used = 0.119

Using the formula M = n x Mr
M = 0.119 x 23 = 2.737 g ≈ 2.74 g


Example 2:
50 cm³ of ammonia of concentration 0.3 mol dm⁻³ reacts with oxygen to form nitrogen gas and water. Calculate the moles of nitrogen gas formed. 

4NH3 + 3O2   →   2N2 + 6H2O

Convert 50 cm³ to 0.05 dm³
Moles of ammonia used = concentration x volume
n = c x v
n = 0.05 x 0.3 = 0.015 mol

Molar ratio of Ammonia : Nitrogen gas = (4 : 2) = (2 : 1)
2 moles of ammonia forms 1 moles of nitrogen gas. So moles of nitrogen gas produced would be half of moles of ammonia produced. 

So moles of nitrogen gas produced = ½ x 0.015 = 0.0075 mol