Question

8c) Consider the reaction represented by the following equation: Au(s)+3Cl2(g)→2AuCl3(s). If 1.250 g of Au and 1.744 g of Cl2 were mixed: (i) determine which of the reactants is in excess; (ii) calculate the excess amount. [Au=197.0, Cl=35.5]

Accepted Answer

Mole of Au = 1.25/197 = 0.006345 mol. Mole of Cl2 = 1.744/71 = 0.0256 mol. From the balanced equation, 2 moles of Au require 3 moles of Cl2 (i.e. 2/2 x 0.006345 mol of Au require 3/2 x0.006345 mol of Cl2) = 0.009518 mol of Cl2. Since the amount of Cl2 given (0.02456 mol) is greater than the amount required (0.009518 mol), Cl2 is the excess reagent. Excess amount = (0.02456-0.009518) mol of Cl2 = 0.01594 mol of Cl2.

WAEC Chemistry 2010 Theory — Question 46

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