Answer:
[tex]\large \boxed{\text{29.2 g}}[/tex]
Explanation:
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
MM: 52.00 151.99
2Cr + … ⟶ Cr₂O₃ + …
m/g: 20.0
(a) Moles of Cr
[tex]\text{Moles of Cr} = \text{20.0 g Cr}\times \dfrac{\text{1 mol Cr }}{\text{52.00 g Cr}}= \text{0.3846 mol Cr}[/tex]
(b) Moles of Cr₂O₃
[tex]\text{Moles of Cr$_{2}$O}_{3} = \text{0.3846 mol Cr} \times \dfrac{\text{1 mol Cr$_{2}$O}_{3}}{\text{2 mol Cr}} = \text{0.1923 mol Cr$_{2}$O}_{3}[/tex]
(c) Mass of Cr₂O₃
[tex]\text{Mass of Cr$_{2}$O}_{3} =\text{0.1923 mol Cr$_{2}$O}_{3} \times \dfrac{\text{151.99 g Cr$_{2}$O}_{3}}{\text{1 mol Cr$_{2}$O}_{3}} = \textbf{29.2 g Cr$_{2}$O}_{3}\\\\\text{The reaction will produce $\large \boxed{\textbf{29.2 g}}$ of Cr$_{2}$O$_{3}$}[/tex]
