Answer:
9.43 L Br₂ (g)
Explanation:
First, write a balanced reaction equation. Then use dimensional analysis to convert from grams of aluminum chloride to moles of aluminum chloride (with molar mass from the periodic table), from moles of aluminum chloride to moles of bromine gas (using the balanced equation), and from moles of bromine gas to liters of bromine gas (remember at STP, 1 mol = 22.4 L). See the attached image for the work out. Finally, account for sig figs to get 9.43 L of bromine gas
