Answer to Question #317822 in General Chemistry for lyndey

Question

Since iron is more reactive than copper, it displaces the copper ions from the aqueous solution of copper sulphate. This is called a single displacement reaction of one metal by another metal. To experimentally test this, Joel reacted 0.9845 g of iron fillings with an excess of copper sulphate solution. When the reaction was complete the solution had changed from blue to green and a copper deposit weighing 1.024 g had formed in the solution. In this reaction, metallic iron is converted into ferrous ion (Fe2+) and cupric ion (Cu2+) is converted into metallic copper.

a) Calculate the number of moles of iron used and the number of moles of copper formed.

b) Calculate the theoretical mass of copper that was expected to be produced and hence determine the percentage yield of Joel’s experiment.

How many molecules of iron are present in 0.9845 g of iron fillings?

Answer 1

Mass of iron used in the reaction = 0.9845 g Molar mass of iron = 55.845 g/mol Hence, we can calculate the moles of iron used by dividing the mass of iron with the atomic mass. Mass of copper produced = 1.024 g Molar mass of copper = 63.546 g/mol