In the given reaction:
$2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}$
How many moles of lead nitrate are needed to produce 2.24 litres of oxygen at STP?

Open in App

Solution

Since the reaction is balanced,
$22.4 L$ of oxygen is produced from 2 mol of lead nitrate.
So, $2.24 L$ will be produced from $(\frac{2}{22.4} \times 2.24) = 0.2$ mol of lead nitrate.

Suggest Corrections

Similar questions

2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}

2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}

(P b = 207; N = 14; O = 16)

2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}

2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}

2 P b {(N O_{3})}_{2 (s)}

H e a t ⟶

2 P b O_{(s)} + 4 N O_{2 (g)} ↑ + O_{2 (g)}

Join BYJU'S Learning Program