ANSWER
99 grams of bromine gas will give us 27.75L
STEP-BY-STEP EXPLANATION:
What to find? Volume of bromine
Given parameters
• The mass of bromine = 99 grams
,• Molar mass of bromine = 79.904 g/mol
Firstly, we need to find the mole of bromine
[tex]\begin{gathered} \text{Mole = }\frac{\text{ reacting mass}}{\text{molar mass}} \\ \text{Mole = }\frac{99}{79.904} \\ \text{Mole = 1.239 moles} \end{gathered}[/tex]Note that, 1 mole = 22.4L at STP
[tex]\begin{gathered} 1\text{ mole }\rightarrow\text{ 22.4L} \\ 1.239\text{ moles }\rightarrow\text{ xL} \\ \text{Cross multiply} \\ xL\cdot\text{ 1 mole = 22.4L }\cdot\text{ 1.239 moles} \\ x\text{ = }\frac{22.4L\cdot\text{ 1.239 moles}}{1\text{ mole}} \\ x\text{ = 27.75L} \end{gathered}[/tex]Hence, 99 grams of bromine gas will give us 27.75L