Consider the following unbalanced equation:
Li2S(aq) + Bi(NO3)3(aq) → LiNO3(aq) + Bi2S3(s)
If 20.5 moles of Li2S(aq) reacts with an excess of Bi(NO3)3(aq), what is the theoretical yield of LiNO3(aq) in moles?

41 moles

Explanation:

The balanced equation for the reaction is given below:

3Li2S + 2Bi(NO3)3 —> 6LiNO3 + Bi2S3

The theoretical yield of LiNO3 can be obtained as follow:

From the balanced equation,

3 moles of Li2S produced 6 moles of LiNO3.

Therefore, 20.5 moles of Li2S will produce = (20.5 x 6)/3 = 41 moles of LiNO3

Therefore, the theoretical yield of LiNO3 is 41 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%