What mass of barium sulfate can be produced when 100.0mL of a .100M solution of barium chloride is mixed with 100mL of a .100M solution of iron(III) sulfate?
2007-04-10 17:02:47 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
3BaCl2 (aq) + Fe2(SO4)3 --> 3BaSO4 (s) + 2FeCl3 (aq)
No. of moles of BaCl2 = 100/1000 x 0.100 = 0.01 mol
No. of moles of Fe2(SO4)3 = 100/1000 x 0.100 = 0.01 mol
Hence, BaCl2 is the limiting reagent.
No. of moles of BaSO4 precipitated = 0.01 mol
Mass of BaSO4 precipitated = 0.01 x (137 + 32 + 4x16) = 2.33 g
2007-04-10 17:10:25 · answer #1 · answered by estheryltan 3 · 0⤊ 0⤋
There are .01 moles of Ba2+ and .01 moles of SO4(2-)
Thus .01 moles of BaSO4 will be formed
Molar mass of BaSO4 = 137+32+64 = 233 g
Thus mass = .01x233 g = 2.33 g
2007-04-11 00:14:08 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋