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

How many grams of barium chloride would actually be necessary to complete the reaction of the silver nitrate?

4.21 g of AgCl

3.06 g of BaCl₂ will be needed to complete the reaction

Explanation:

The first step is to determine the reaction.

Reactants: BaCl₂ and AgNO₃

The products will be the silver chloride (AgCl) and the Ba(NO₃)₂

The reaction is: BaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) ↓ + Ba(NO₃)₂ (aq)

We determine the silver nitrate moles: 5 g . 1mol / 169.87 g = 0.0294 moles. Now, according to stoichiometry, we know that ratio is 2:2-

2 moles of nitrate can produce 2 moles of chloride, so the 0.0294 moles of silver nitrate, will produce the same amount of chloride.

We convert the moles to mass → 143.32 g / mol . 0.0294 mol = 4.21 g of AgCl.

Now, we consider the BaCl₂.

2 moles of nitrate can react to 1 mol of barium chloride

Then, 0.0294 moles of silver nitrate will react to (0.0294 . 1) /2 = 0.0147 moles. We convert the moles to mass:

0.0147 mol . 208.23 g /1mol = 3.06 g of BaCl₂

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