Question

PLEASE ANSWER IT'S URGENT 40 POINTS

2C6H6(g) + 150₂(g) → 12CO₂(g) + 6H₂O(g)
What volume of water vapor, in liters,
forms at STP when 1 L of C6H6 reacts
with oxygen?
[?] LH₂O
Volume (L) H₂O
Enter

Answer 1

Volume of 1 mole of a gas at STP =22.4L

Volume of 15 mole of a gas at STP =22.4×15=336L

Molecular weight of benzene =78g

No. of moles of benzene in 39g of benzene =

78

39

=0.5 mole

From the given reaction,

Volume of oxygen required to burn 2 mole of benzene =336L.

Therefore,

Volume of oxygen required to burn 0.5 mole of benzene =

2

336

×0.5=84L

Hence 84 litres of O

2

at STP are needed to complete the combustion of 39g of liquid benzene.