Answer:
[tex]\large \boxed{\textbf{32.7 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: 102.89 159.81
2NaBr + F₂ ⟶ 2NaF + Br₂
m/g: 25.4
(a) Moles of Br₂
[tex]\text{Moles of Br$_{2}$} = \text{25.4 g Br$_{2}$}\times \dfrac{\text{1 mol Br$_{2}$}}{\text{159.81 g Br$_{2}$}}= \text{0.1589 mol Br$_{2}$}[/tex]
(b) Moles of NaBr
[tex]\text{Moles of NaBr} = \text{0.1589 mol Br$_{2}$} \times \dfrac{\text{2 mol NaBr }}{\text{1 mol Br$_{2}$}} = \text{0.3179 mol NaBr}[/tex]
(c) Mass of NaBr
[tex]\text{Mass of NaBr} =\text{0.3179 mol NaBr} \times \dfrac{\text{102.89 g NaBr}}{\text{1 mol NaBr }} = \textbf{32.7g NaBr}\\\\\text{The reaction produces $\large \boxed{\textbf{32.7 g}}$ of NaBr}[/tex]
