Propane burns as $C X_{3} H X_{8} + 5 O X_{2} 3 CO X_{2} + 4 H X_{2} O$ . Air contains $21%$ $O X_{2}$ by volume. What volume of air at STP (in litres, nearest integer) is needed to burn $6.6 g$ of propane completely? (C = 12, H = 1; $22.4 L mo l^{- 1}$ )

Numerical answer type, answer in L, tolerance 0.5.

Answer: 80 L

n (C X_{3} H X_{8}) = 6.6/44 = 0.15

mol, needing

0.75

mol

O X_{2}

, i.e.

16.8 L

at STP. Air

= 16.8/0.21 = 80 L