Respuesta :
Answer:
Based on the given information, the balanced equation is:
NaCl + AgNO3 = AgCl + NaNO3
Now the moles present in 50 ml of 1M NaCl is,
= 50 * 1/1000 mole = 0.05 mole
And the moles present in 50 ml of 1 M AgNO3 is,
= 50*1/1000 mole = 0.05 mole
Therefore, 0.05 mole of NaCl combines with 0.05 mole of AgNO3 to precipitate 0.05 mole of AgCl.
Now in the second case, to balance the chemical equation, 50 ml of 2 M NaCl combines with 50 ml of 1 M AgNO3. So, the moles present in 50 ml of 2 M NaCl will be,
= 50*2/1000 mole = 0.1 mole
However, the amount of AgNO3 in the second case is not changing, therefore, the amount of AgCl precipitated will be,
= 0.1 - 0.05 = 0.05 mole
Therefore, the amount of precipitate would not change as there is no change in the amount of AgNO3.
The amount of precipitate (NaCl) formed would be 0.05 mole
From the question,
We are to determine the amount of precipitate formed.
First, we will write the balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)
This means
1 mole of NaCl reacts with 1 mole of AgNO₃ to produce 1 mole of AgCl and 1 mole of NaNO₃
The precipitate formed is AgCl
Now, we will determine the number of moles of each reactant present
- For NaCl
Volume = 50.0 mL = 0.05 L
Concentration = 2.0 M
Using the formula
Number of moles = Concentration × Volume
∴ Number of moles of NaCl present = 2.0 × 0.05
Number of moles of NaCl present = 0.10 mole
- For AgNO₃
Volume = 50.0 mL = 0.05 L
Concentration = 1.0 M
∴ Number of moles of AgNO₃present = 1.0 × 0.05
Number of moles of AgNO₃ present = 0.05 mole
Since
1 mole of NaCl reacts with 1 mole of AgNO₃ to produce 1 mole of AgCl
Then,
0.05 mole of NaCl will react with 0.05 mole of AgNO₃ to produce 0.05 mole of AgCl
∴ The number of moles of NaCl formed is 0.05 mole
Hence, the amount of precipitate (NaCl) formed would be 0.05 mole
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