If 19.10 grams of zinc react with 21.6 grams of iron (iii) chloride, how many grams of iron is prouced? How many grams of excess reactant remains?

Answer 1

First, balance the reaction: 3 Zn + 2 FeCl₃ --> 3 ZnCl₂ + 2 Fe.

(19.10g Zn) * (1 mole Zn /65.38g Zn) * (2 mole Fe /3 mole Zn) = 0.1947₆ mole Fe.

(21.6g FeCl₃) * (1 mole FeCl₃ /162.2g FeCl₃) * (2 mole Fe /2 mole FeCl₃) = 0.133₁₇ mole Fe.

Iron (III) chloride is limiting, so mass of iron is (0.133₁₇ mole Fe) * (55.845g Fe /1 mole Fe) = 7.44g Fe.

Excess (Δmole Fe = 0.061₅₉ mole Fe) * (3 mole Zn /2 mol Fe) * (65.38g Zn /1 mole Zn) = 6.0₄ g Zn remaining (only 2 sig figs here).