How much sodium is needed to interact with 11.2 liters of oxygen?

Sodium can interact with oxygen in different ways. Depending on the conditions, Na2O oxide, Na2O2 peroxide or NaO2 superoxide can be formed. Let us assume that the reaction proceeds with the formation of an oxide.

4Na + O2 = 2Na2O

Let’s find the amount of oxygen substance:

n (O2) = V (O2) / VM = 11.2 / 22.4 = 0.5 mol;

According to the stoichiometry of the reaction:

n (Na) = 0.25 n (O2) = 0.5 * 0.25 = 0.125 mol;

Find the mass of sodium:

m (Na) = n (Na) * M (Na) = 0.125 * 23 = 2.875 g.

Answer: m (Na) = 2.875 g.

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