How do you do conversion problems?

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Ответ:

sharondacarruth1656

16.12.2020

Chemistry

$$\large \boxed{\text{20.6 g}}$

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

$\text{Moles of LiClO}_{3} = \text{38.7 g LiClO}_{3}\times \dfrac{\text{1 mol LiClO}_{3}}{\text{90.39 g LiClO}_{3}}= \text{0.4281 mol LiClO}_{3}$

(b) Moles of O₂

$\text{Moles of O$_{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}$}$

$\text{Mass of O$_{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}$

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