Barium chloride was used to precipitate silver chloride from a solution of silver nitrate.What mass of barium chloride had to react if 0.635 g of silver chloride formed?

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction is:

AgNO3 + BaCl2 → AgCl + Ba(NO3)2

From the balanced equation, we can see that 1 mole of BaCl2 reacts with 1 mole of AgCl.

The molar mass of AgCl is 143.32 g/mol, so 0.635 g of AgCl is equal to:

0.635 g / 143.32 g/mol = 0.00442 mol AgCl

Since 1 mole of BaCl2 reacts with 1 mole of AgCl, we need 0.00442 mol of BaCl2 to react with the AgNO3. The molar mass of BaCl2 is 208.23 g/mol, so the mass of BaCl2 needed is:

0.00442 mol x 208.23 g/mol = 0.921 g

Rounding to three significant figures, the answer is 0.461 g, which is choice B.

Barium Chloride Was Used to Precipitate Silver Chloride from a Solution