Answer:
[tex]$\large \boxed{\text{20.6 g}}[/tex]
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
MM: 90.39 32.00
2LiClO₃ ⟶ 2LiCl + 3O₂
m/g: 38.7
(a) Moles of LiClO₃
[tex]\text{Moles of LiClO}_{3} = \text{38.7 g LiClO}_{3}\times \dfrac{\text{1 mol LiClO}_{3}}{\text{90.39 g LiClO}_{3}}= \text{0.4281 mol LiClO}_{3}[/tex]
(b) Moles of O₂
[tex]\text{Moles of O$_{2}$} = \text{0.4281 mol LiClO}_{3} \times \dfrac{\text{3 mol O$_{2}$}}{\text{2 mol LiClO}_{3}} = \text{0.6422 mol O$_{2}$}[/tex]
(c) Mass of O₂
[tex]\text{Mass of O$_{2}$} =\text{0.6422 mol O$_{2}$} \times \dfrac{\text{32.00 g O$_{2}$}}{\text{1 mol O$_{2}$}} = \textbf{20.6 g O$_{2}$}\\\\\text{The reaction can produce $\large \boxed{\textbf{20.6 g}}$ of oxygen}[/tex]
