When silver nitrate reacts with barium chloride, silver chloride and barium nitrate are formed. how many grams of silver chloride are formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride? when silver nitrate reacts with barium chloride, silver chloride and barium nitrate are formed. how many grams of silver chloride are formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride? 18.8 g 8.44 g 11.9 g 9.40 g

Answer : The mass of silver chloride formed are, 8.44 grams.

Explanation : Given,

Mass of = 10.0 g

Mass of = 15.0 g

Molar mass of = 169.87 g/mole

Molar mass of = 208.23 g/mole

Molar mass of = 143.32 g/mole

First we have to calculate the moles of and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 moles of react with 1 mole of

So, 0.0589 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of .

As, 2 moles of react to give 2 moles of

So, 0.059 moles of react to give 0.059 moles of

Now we have to calculate the mass of .

Therefore, the mass of silver chloride formed are, 8.44 grams.

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