The number of moles of silver chloride that will precipitate is 0.008 mole
From the question,
We are to determine the number of moles of silver chloride that will precipitate
First,
We will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
NaCl + AgNO₃ → AgCl + NaNO₃
This means
1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride and 1 mole of sodium nitrate
Now, we will determine the number of moles of each reactant present
- For sodium chloride (NaCl)
Concentration = 0.440 M
Volume = 25.0 mL = 0.025 L
Using the formula
Number of moles = Concentration × Volume
∴ Number of moles of NaCl present = 0.440 × 0.025
Number of moles of NaCl present = 0.011 mole
- For silver nitrate (NaNO₃)
Concentration = 0.320 M
Volume = 25.0 mL = 0.025 L
∴ Number of moles of NaNO₃ present = 0.320 × 0.025
Number of moles of NaNO₃ present = 0.008 mole
From the balanced chemical equation,
1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride
Then,
0.008 mole of sodium chloride will react with the 0.008 mole of silver nitrate to produce 0.008 mole of silver chloride
∴ 0.008 mole of silver chloride will be produced
Hence, the number of moles of silver chloride that will precipitate is 0.008 mole
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