How many grams of product might form for the following reaction if 33.8 L of Oxygen gas is used in the following reaction? LiCl + O2 -> LiClO3

91.41 g of LiClO₃.

Explanation:

We'll begin by calculating the number of mole of O₂ that occupied 33.8 L. This can be obtained as follow:

22.4 L = 1 mole of O₂

Therefore,

33.8 L = 33.8 L × 1 mole / 22.4 L

33.8 L = 1.51 mole of O₂

Next, the balanced equation for the reaction.

2LiCl + 3O₂ —> 2LiClO₃

From the balanced equation above,

3 moles of O₂ reacted to produce 2 moles of LiClO₃.

Therefore, 1.51 mole of O₂ will react to produce = (1.51 × 2)/3 = 1.01 mole of LiClO₃.

Finally, we shall determine the mass of 1.01 mole of LiClO₃. This can be obtained as follow:

Mole of LiClO₃ = 1.01 mole

Molar mass of LiClO₃ = 7 + 35.5 + (3×16)

= 7 + 35.5 + 48

= 90.5 g/mol

Mass of LiClO₃ =?

Mass = mole × molar mass

Mass of LiClO₃ = 1.01 × 90.5

Mass of LiClO₃ = 91.41 g

Thus, 91.41 g of LiClO₃ were obtained from the reaction.