Question

Phosphorous burns in oxygen according to the equation
P4 + 5O2 P4O . How many litres of oxygen will be required at s.t.p. for complete oxidation
of 12.4g of phosphorous? (P = 31, 0 = 16 and molar volume of a gas at s.t.p. = 22.4 litres)

Answer 1

Molar mass of $P_4=31×4=124$

Molar mass of $O_2=16×2=32$

Moles of $P_4=\frac{12.4}{124}=0.1moles$

From equation

$P_4 + 5O_2----> P_4O .$

Moles of $O_2= 5×0.1= 0.5moles$

Litres of oxygen = $0.5×22.4=11.2litres$