After reacting all of the #FeS# in a #1.5\ L# of a reaction solution that had #[FeS]=0.2\ M#, how many grams of #Fe_2O_3# were formed? #" "# # # #4FeS+7O_2\rightarrow 2Fe_2O_3+4SO_2#
1 Answer
Apr 2, 2018
Explanation:
It’s given that
#["FeS"] = "0.2 M"# #"Volume of solution = 1.5 L"#
#"Molarity" = "Moles of solute"/"Volume of solution (in litres)"#
So, we have
#underbrace("4FeS") + "7O"_2 -> underbrace("2Fe"_2"O"_3) + "4SO"_2#
#color(blue)("4 mol") color(white)(................)color(blue)("2 mol")#
From above equation,
We have
#0.3 cancel"mol FeS" × ("2 mol Fe"_2"O"_3)/(4 cancel"mol FeS") = "0.15 mol Fe"_2"O"_3#
Molar mass of
Mass of