Question

2. How many grams of sodium bromide must be reacted with excess flourine if 25.4 g of bromine gas is produced? 2 NaBr(aq) + F2(g) → 2 NaF(aq) + Br2(l)

Answer 1

`\large \boxed{\textbf{32.7 g}}`

Step-by-step 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

`\text{Moles of Br$_(2)$} = \text{25.4 g Br$_(2)$}* \frac{\text{1 mol Br$_(2)$}}{\text{159.81 g Br$_(2)$}}= \text{0.1589 mol Br$_(2)$}`

(b) Moles of NaBr

`\text{Moles of NaBr} = \text{0.1589 mol Br$_(2)$} * \frac{\text{2 mol NaBr }}{\text{1 mol Br$_(2)$}} = \text{0.3179 mol NaBr}`

(c) Mass of NaBr

`\text{Mass of NaBr} =\text{0.3179 mol NaBr} * \frac{\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}`