Answer:
91.41 g of LiClO₃.
Step-by-step 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.