Answer:
85.99 grams of AgCl will be formed.
Step-by-step explanation:
To determine the grams of AgCl formed in the reaction, we need to consider the stoichiometry of the reaction between silver nitrate (AgNO3) and barium chloride (BaCl2):
AgNO3 + BaCl2 -> AgCl + Ba(NO3)2
The balanced equation shows that the molar ratio between AgNO3 and AgCl is 1:1. This means that 1 mole of AgNO3 produces 1 mole of AgCl.
Given:
Volume of BaCl2 solution = 1500 ml = 1.5 L
Molarity of BaCl2 solution = 0.400 M
First, we need to calculate the number of moles of BaCl2 present in the solution:
moles of BaCl2 = volume of BaCl2 solution * molarity of BaCl2 solution
= 1.5 L * 0.400 M
= 0.600 moles
Since the molar ratio between BaCl2 and AgNO3 is 1:1, the number of moles of AgNO3 needed for complete reaction is also 0.600 moles.
Now, using the molar mass of AgCl, which is 143.32 g/mol, we can calculate the grams of AgCl formed:
grams of AgCl = moles of AgNO3 * molar mass of AgCl
= 0.600 moles * 143.32 g/mol
= 85.99 grams
Therefore, by adding enough AgNO3 to react fully with the 1500 ml of 0.400 M BaCl2 solution, approximately 85.99 grams of AgCl will be formed.