What volume of air is required for the interaction of 17.5 g of lithium containing 20% impurities?

What volume of air is required for the interaction of 17.5 g of lithium containing 20% impurities? The reaction produces lithium oxide.

1) 2Li + O₂ = Li₂O

2) w% (Li) = 100% -20% = 80%

3) m (Li) = 17.5 * 80/100 = 14g

4) n (Li) = 14/7 = 2 mol
5) n (O₂) = 2/2 = 1 mol, according to the equation.
6) in 100 liters of air. solder 21 l О₂

in xl air. contains 22.4 l of О₂

x = 106.7 l

Answer: 106.7 l