Question

For the following reaction, 21.7 grams of iron are allowed to react with 22.1 grams of hydrochloric acid iron(s) + hydrochloric acid (aq) — iron(II) chloride(aq) + hydrogen (g) What is the maximum amount of iron(II) chloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Group 1 more group attempt remaining

Answer 1

21-7 gm Fe molar mass of fe= 56 g/mol x moles of pe= 21.77 = 0:39.modes 22-1 gm . Hal molar mass of Hd = 36-59 mal a moles of HI = 22.1 = 0.605 moles 36.5 5 3 Fe + 2HCl - fecha + H2 Dolg mol Oocos me". . 2 mol Hill need 1 mob.re so .0.605 mol Hol will need / 0-60s = 0. 303 mol Fe and we have enough fe so fe is excess reactant and Hel is limithing reactant. 2 mol Hal gives I mol fech so 0.605 mol Hel will gine 0.303 mol feal m olar mass of fedz = 177 gimol a marinimass of feck produced. = 2.303X127 38-5 gm - 3 7 å limitting reagent = Hee fe reacted = 0.303 moles fe left = 0.39-0-303 = 0-087 moles a mass of fe (excess stea ct avet) left = 02087x56 1 Scanned with CamScanner = 4.9gm