Q. Example on Law of Conservation of Mass?

A slim strip of iron with a mass of 15.72 g is placed in to a solution containing 21.12 g of copper and copper (II) sulphate begins to form. After a short time the reaction stops because all of the copper (II) sulphate has reacted. The iron strip is acquired to have a mass of 8.33 g The mass of copper formed is found to be 8.41 g and What mass of iron (II) sulphate has been formed in the reaction?

Solution concerns the law of conservation of mass. In this chemical reaction there are two reactants and two products thus the law of conservation of mass can be restated as follows.

Mass_{reactant 1} +  Mass_{reactant 2} =  Mass_{product 1}+ Mass_{product 2}

Modify the equation with the names of the reactants and products.

Mass_iron + Mass_{copper sulfate} = Mass_copper + Mass_{iron sulfate}

To discover the mass of iron sulfate rearrange the equation.

Mass_{iron sulfate}= Mass_iron  + Mass_{copper sulfate}- Mass_copper

Then find out the mass of iron that reacted.

Massiron = original mass of iron - mass of iron remaining

Massiron = 15.72 g - 8.33 g = 7.39 g

At last substitute the masses into the equation and solve.

Mass_{iron sulfate}= 7.39 g + 21.12 g - 8.41 g = 20.10 g iron sulfate

To verify your work make sure the sum of the masses of the reactants is equal to the sum of the masses of the products.