Assume we want to use 2 g of Ba(NO3)2 in an experiment that involves a reaction with CuSO4. How much CuSO4 is required and how much BaSO4 can we get?
When barium nitrate Ba(NO3)2 reacts with CuSO4, the equation of the chemical reaction is Ba(NO3)2 + CuSO4 --> BaSO4 + Cu(NO3)2.
It can be seen that each atom of barium nitrate requires one atom of copper sulfate and the products contains one atom of barium sulfate.
Barium nitrate has a molar mass of 261.37 g/mol, CuSO4 has a molar mass of 159.62 g/mol and BaSO4 has a molar mass of 233.43 g/mol.
2 grams of barium nitrate is equal to 2/261.37 moles. The reaction requires 2/261.37 moles of copper sulfate which weighs (2/261.37)*159.62 = 1.22 g
The product consists of 2/261.37 moles of barium sulfate. This is equivalent to 1.78 g.
The given reaction requires 1.22 g of CuSO4 and the product consists of 1.78 g of BaSO4.