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Ok here goes! moles of Cl2= 2/71 = 0.028 moles of Fe = 2/3 x 0.028 = 0.0187 mass of Fe = 0.0187 x 57 = 1.064g or other way moles of Fe = 0.018 moles of l2 = 0.027 mass of Cl2 = 1.917g therefore Cl2 in excess. mass of reactant in excess left over = 2.0 - 1.917 = 0.083g Thanks

Thanks Tiny Tim. So if I did the following am I on the right track? Moles of Fe = 1/112 and moles of Cl2 = 2/213 (taking into account stoiciometry). Therefore Cl2 in excess. Mass of Cl2 left after reaction: 9.39x10-3 - 8.92x10-3 = 4.7x10-4 Mass = 4.7x10-4 x 71 = 0.0334g Any good?

1.0 g of Iron metal reacts with 2.0 g of chlorine gas to make iron (III) chloride, FeCl3. (a) Write a balanced equation for this reaction. (b) Which of the reactants is present in excess ? (c) What mass of this reactant will be left over at the end of the reaction ? (7 marks) Homework...