Respuesta :
Answer:
The amount of precipitate formed would 7.175 grams of silver chloride.
Explanation:
[tex]Moles (n)=Molarity(M)\times Volume (L)[/tex]
Moles of NaCl = n
Volume of NaCl solution = 50.0 mL = 0.050 L
Molarity of the hydrogen peroxide = 2.0 M
[tex]n=2.0 M\times 0.050 L=0.100 mol[/tex]
Moles of silver nitarte = n'
Volume of silver nitrate solution = 50.0 mL = 0.050 L
Molarity of the silver nitrate = 1.0 M
[tex]n'=1.0 M\times 0.050 L=0.050 mol[/tex]
[tex]NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)[/tex]
According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :
[tex]\frac{1}{1}\times 0.050 mol=0.050 mol[/tex] 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;
[tex]\frac{1}{1}\times 0.050 mol=0.050 mol[/tex] of AgCl
Mass of 0.050 moles of AgCl ;
[tex]0.050 mol\times 143.5 g/mol=7.175 g[/tex]
The amount of precipitate formed would 7.175 grams of silver chloride.
The amount of sodium nitrate precipitate formed has been 7.175 grams.
The moles of a compound has been given by :
Molarity = [tex]\rm \dfrac{Moles}{Volume}[/tex]
Moles = Molarity × Volume
- The moles of NaCl has been:
Moles of NaCl = 0.05 L × 2 M
Moles of NaCl = 0.1 mol
- Moles of Silver nitrate has been:
Moles of [tex]\rm AgNO_3[/tex] = 0.05 L × 1 M
Moles of [tex]\rm AgNO_3[/tex] = 0.05 mol
The balanced chemical reaction has been:
[tex]\rm NaCl\;+\;AgNO_3\;\rightarrow\;AgCl\;+\;NaNO_3[/tex]
The precipitate has been sodium nitrate, the reaction states that:
1 mole NaCl requires 1 mole of silver nitrate, thus 0.1 mol NaCl requires, 0.1 mol silver nitrate. Since there has been lesser amount of the silver nitrate, thus silver nitrate has been the limiting reactant.
Thus, the moles of sodium nitrate formed has been:
1 mole silver nitrate = 1 mole sodium nitrate
0.05 mol silver nitrate = 0.05 mol sodium nitrate.
The mass of sodium nitrate can be given by:
Mass = Moles × Molecular weight
Mass of sodium nitrate = 0.05 mol × 143.5 g/mol
Mass of sodium nitrate = 7.175 grams.
The amount of sodium nitrate precipitate formed has been 7.175 grams.
For more information about the balanced equation, refer to the link:
https://brainly.com/question/8062886
