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Answer:
The amount of precipitate formed would 7.175 grams of silver chloride.
Explanation:
Moles of NaCl = n
Volume of NaCl solution = 50.0 mL = 0.050 L
Molarity of the hydrogen peroxide = 2.0 M
Moles of silver nitarte = n'
Volume of silver nitrate solution = 50.0 mL = 0.050 L
Molarity of the silver nitrate = 1.0 M
According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :
of NaCl
This means that silver nitrate is in limiting amount and NaCl is in excessive amount.
So, the amount of AgCl depends upon amount of silver nitrate.
According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.
Then 0.050 moles of silver nitrate will give;
of AgCl
Mass of 0.050 moles of AgCl ;
The amount of precipitate formed would 7.175 grams of silver chloride.