Answer: 1.43 g.
Step-by-step explanation:
When silver nitrate and calcium chloride are mixed, they react to form a white precipitate of silver chloride, according to the following balanced chemical equation:
AgNO3 + CaCl2 → AgCl + Ca(NO3)2
The equation shows that one mole of silver chloride is produced for every one mole of silver nitrate used. Therefore, we can find the moles of silver chloride produced by calculating the moles of silver nitrate used in the reaction.
First, we can calculate the moles of silver nitrate used:
moles of AgNO3 = volume of AgNO3 x concentration of AgNO3
= 50.0 ml x 0.200 mol/L
= 0.0100 mol
Since the stoichiometry of the reaction is 1:1, we know that 0.0100 moles of silver chloride will be produced.
The molar mass of silver chloride is 143.32 g/mol. Therefore, the mass of the silver chloride precipitate can be calculated as:
mass of AgCl = moles of AgCl x molar mass of AgCl
= 0.0100 mol x 143.32 g/mol
= 1.43 g
Therefore, the mass of the silver chloride precipitate formed is 1.43 g.