Question

Use the References to access important values If needed for this question. For the following reaction, 23.5 grams of iron are allowed to react with 24.3 grams of hydrochloric acid. Iron (8) + hydrochloric acid (aq) — iron(II) chloride (aq) + hydrogen () What is the maximum amount of iron(II) chloride that can be formed? What is the FORMULA for the limiting reagent? grams What amount of the excess reagent remains after the reaction is complete? Submit Answer Retry Entire Group 7 more group attempts remaining

Answer 1

Answer

maximum amount of FeCl2 than be formed = 42.24g

formula of limiting reagent = HCl

amount of excess reagent after the reaction is complete = 4.89g

Explanation

Fe(s) + 2HCl(aq) -------> FeCl2(aq) + H2(g)

soichiometrically, 1mole of Fe react with 2moles of HCl

given moles of Fe = 23.5g/55.85g/mol = 0.4208mol

given moles of HCl = 24.3g/36.46g/mol = 0.6665mol

0.4208moles of Fe react with 0.8416moles of HCl but available moles of HCl is 0.6665mol ,so

Limiting reagent is HCl

stoichiometrically, 2moles of HCl gives 1moles of FeCl2

moles of FeCl2 obtained by 0.6665moles of HCl = 0.6665mol/2 = 0.33325mol

maximum amount of FeCl2 that can be formed = 0.33325mol × 126.75g/mol=42.24g

number of moles of Fe react with 0.6665moles of HCl= 0.6665mol/2 = 0.33325mol

mass of Fe reacted = 0.33325mol × 55.85g/mol = 18.61g

mass of Fe remaining after reaction = 23.5g - 18.61g = 4.89g