Question

how many litre of oxygen at stp is required to burn 60g C2H6

Answer 1

2C₂H₆+7O₂---aCO₂+6H₂O

7*22.4lit of O₂=    156.8 lit of O₂

Answer 2

Answer: 156.8 L

Explanation:-

$2C_2H_6(g)+7O2(g)\rightarrow 4CO_2(g)+6H_2O(l)$

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

given mass of ethane = 60 g

Molar mass of ethane = 60g/mol

$Moles=\frac{60g}{30g/mol}=2moles$

From the given balanced chemical equation:

2 moles of ethane require 7 moles of oxygen.

According to Avogadro's law, 1 mole of every gas occupies 22.4 L at STP.

Thus 2 moles of ethane require=$7\times 22.4=156.8L$ of oxygen.