Question

Given 20.0 of Cr, how many grams of the product Cr2O3 could be produced?

Answer 1

`\large \boxed{\text{29.2 g}}`

Step-by-step 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

`\text{Moles of Cr} = \text{20.0 g Cr}* \frac{\text{1 mol Cr }}{\text{52.00 g Cr}}= \text{0.3846 mol Cr}`

(b) Moles of Cr₂O₃

`\text{Moles of Cr$_(2)$O}_(3) = \text{0.3846 mol Cr} * \frac{\text{1 mol Cr$_(2)$O}_(3)}{\text{2 mol Cr}} = \text{0.1923 mol Cr$_(2)$O}_(3)`

(c) Mass of Cr₂O₃

`\text{Mass of Cr$_(2)$O}_(3) =\text{0.1923 mol Cr$_(2)$O}_(3) * \frac{\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$_(3)$}`