How many grams of silver cholride will be precipitated by adding sufficient silver nitrate to react with 1500.0ml of.400m barium chloride solution?

172.2 g of is formed by adding sufficient silver nitrate to react with 1500.0mL of 0.400M barium chloride solution.

Explanation:

To calculate the moles, we use the equation:

According to stoichiometry:

1 mole of produce = 2 moles of

Thus 0.6 moles will produce = moles of

Mass of

Thus 172.2 g of is formed by adding sufficient silver nitrate to react with 1500.0mL of 0.400M barium chloride solution.

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