5. identifying the limiting reagent: given the reaction 2cu + s → cu2s, if you had 3 moles of both cu and s, which would you run out of first?

The reaction is

2Cu + S → Cu2S

So as per balanced given reaction we need one mole of sulphur over two moles of copper

Or

To react with one mole of sulphur we need two moles of copper

Thus we can say that the mole ration required for complete reaction will be

Cu : S = 2 : 1

As per the given problem the moles of each Cu and S are 3 moles

Thus for each three moles of given sulphur we need six moles of Cu to give three moles of Cu2S

However we have only three moles of copper available

So the limiting reagent is copper

It is going to decide the actual yield of reaction

The three moles of Copper will get completely utilized and only 1.5 moles of S will be used in the reaction making S an excess reagent

The moles of S left will be = 1.5 moles

moles of Cu2S formed = 1.5 moles

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%