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

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