What is the mass of 2.35 moles of (NH4)3PO4?

The mass of 2.35 moles of (NH4)3PO4 is 411.156 or 411.156 grams/g.

Explanation:

By calculating the mass of the formula, (NH4)3PO4, which equals up to 174.96 grams you divide 2.35 moles by 174.96 grams to equal 411.156 grams. The way to find out the mass is to multiple the molar mass of each chemical to the number beside it such as H multiplied by 4, P multiplied by 3, and so on.

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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%