If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced?

 0.745 g of will be produced from  1.08 g of sodium sulfate

Explanation:

To calculate the moles :

is the limiting reagent as it limits the formation of product and is the excess reagent.

According to stoichiometry :

3 moles of produce = 3 moles of

Thus 0.0076 moles of will require=  of

Mass of

Thus 0.745 g of will be produced from  1.08 g of sodium sulfate

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%