How much air will be required for the interaction of 17 g of magnesium containing 20%

How much air will be required for the interaction of 17 g of magnesium containing 20% of impurities if magnesium oxide is formed as a result of the reaction.

Let’s find the mass of pure magnesium without impurities.

100% – 20% = 80%.

17g – 100%,

X – 80%,

X = (17g × 80%): 100 = 13.6g.

Let’s find the amount of magnesium substance by the formula:

n = m: M.

M (Mg) = 24 g / mol.

n = 13.6 g: 24 g / mol = 0.57 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

2Mg + O2 = 2MgO.

According to the reaction equation, there is 1 mole of oxygen for 2 mol of magnesium. The substances are in quantitative ratios of 2: 1. The amount of oxygen will be 2 times less than the amount of magnesium.

n (О2) = 2n (Mg) = 0.57: 2 = 0.285 mol.

Let’s find the volume of oxygen.

V = Vn n, where Vn is the molar volume of gas equal to 22.4 l / mol, and n is the amount of substance.

V = 0.285 mol × 22.4 L / mol = 6.38 L.

Oxygen in the air is 21%.

6.38 L – 21%,

V (air) – 100%.

V (air) = (6.38 l × 100%): 21% = 30.38 l.

Answer: 30.38 liters.