To obtain iron oxide (11), 2.24 liters of oxygen were consumed. How much iron oxide did you get?

Let’s execute the solution:

According to the condition of the problem, we write down the equation of the process:
V = 2.24 l Xg -?

2Fe + O2 = 2FeO – compounds, iron oxide was obtained.

Calculations according to the formulas of substances:
M (O2) = 32 g / mol;

M (FeO) = 71.8 g / mol.

Proportions:
1 mol of gas at normal level – 22.4 liters;

X mol (O2) – 2.24 L from here, X mol (O2) = 1 * 2.24 / 22.4 = 0.1 mol;

X mol (FeO) – 0.1 mol (O2);

2 mol -1 mol hence, X mol (FeO) = 2 * 0.1 / 1 = 0.2 mol.

Find the mass of the product:
m (FeO) = Y * M = 0.2 * 71.8 = 14.36 g

Answer: the mass of iron oxide (2) is 14.36 g