Question

The volume of O2 gas required at STP to burn completely 29 g of butane (C4H10) gas is
0 74.2 L
56 L
22.4L
0 72.8 L​

Answer 1

Answer:

72.8

Explanation:

116 g of C4H10 gives 13x22.4 L of O2

so... 29 gm of C4H10 gives

{(13x22.4)/116}x29 .... = 72.8 L

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Answer 2

Answer:

The volume of O2 gas required at STP to burn completely 29 g of butane (C4H10) gas is 72.8 L

Explanation:

Chemical equation of Butane combustion is:

2C₄H₁₀ +130₂-------->8CO₂  + 10H₂0

2moles [116 g ] of Butane requires 13 moles of oxygen

As we already know that,

1mole of any gas at STP, occupies 22.4 L

so 13 moles of O2 occupies 13x22.4L

116 g of butane ----------> 13x 22.4 L

29 of butane -----------?

=[(13x22.4)/116 ]x29

=72.8L

∴The volume of O2 gas required at STP to burn completely 29 g of butane (C4H10) gas is 72.8L