Question

Iron combines with 5 g of Copper (1) Nitrate to form 6.01 g of Iron (1) Nitrate and 0.4 g of copper

metal. How much Iron was required to complete this reaction?

Answer 1

First, you would add the products: 0.4 + 6.01 = 6.41 g
Then, to get the mass of Iron required, you would do 6.41 - 5.0 = 1.41
This is because in the conservation of mass principle it states,
mass of reactants = mass of products.