A chemist adds 280.0mL of a 1.6M iron (III) bromide solution to a reaction flask. Calculate the mass in kg of iron (III) bromide the chemist has added to the flask. Round your answer to 2 sig figs. **Transcription for Educational Website:**A chemist adds 280.0 mL of a 1.6 M iron(III) bromide (\( \text{FeBr}_3 \)) solution to a reaction flask. Calculate the mass in kilograms of iron(III) bromide the chemist has added to the flask. Round your answer to 2 significant digits.**Input Box:**- A box labeled "kg" for inputting the answer.**Buttons:**- A button with an "x" icon.- A refresh button.- A question mark button for help.**Options:**- "Explanation" button to view an explanation of the solution.- "Check" button to verify the answer.**Footer:**- Copyright © 2021 McGraw Hill LLC. All Rights Reserved.- Links: Terms of Use, Privacy Center, Accessibility.**Graph/Diagram Explanation:**There are no graphs or diagrams present on this page.

Given :- Molarity of FeBr3 solution = 1.6 M or 1.6 mol/L Volume of solution = 280.0 mL To calculate :- Amount of FeBr3 (in kg) Number of moles of FeBr3 = molarity of FeBr3 solution × volume (L) = 1.6 mol/L × 0.280 L = 0.448 mol Mass of FeBr3 = moles of FeBr3 × molar mass of FeBr3 = 0.448 mol × 295.56 g/mol = 132.4 g Finally convert g to Kg :- 132.4g × (1kg/1000g) = 0.13 Kg. (2 significant figures) Hence, 0.13 Kg of FeBr3 is required.