How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride?

4.21 grams of silver chloride are produced

Explanation:

Let's follow the reaction:

2AgNO₃  +  BaCl₂  →  2AgCl  +  Ba(NO₃)₂

Ratio is 2:2 so 2 moles of nitrate produce, 2 mol of chloride.

Mol = Mass / Molar mass → 5g / 169.87 g/m = 0.0294 moles

So, 0.0294 moles of Silver chloride are produced

Molar mass AgCl = 143.32 g/m

Molar mass . mol = Mass → 143.32 g/m . 0.0294 mol = 4.21 grams

answer: your mm

explanation:

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%