How many moles are in 48.1 grams of FeF3?

I am not sure which number would be considered correct.

unrounded: 5,427.5559

rounded to the nearest thousandth: 5,427.556

rounded to the nearest hundredth: 5,427.56

rounded to the nearest tenth: 5,427.6

rounded to the nearest whole number: 5,428

Explanation:

112.839 * 48.1 = 5,427.5559

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%