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Molar Ratios

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


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