Respuesta :
Answer: 3.6 grams
Explanation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of silver nitrate}=\frac{4.22g}{170g/mol}=0.025moles[/tex]
[tex]\text{Number of moles of aluminium chloride}=\frac{7.73g}{133g/mol}=0.06moles[/tex]
[tex]3AgNO_3(aq)+AlCl_3(aq)\rightarrow Al(NO)_3)_3(aq)+3AgCl(s)[/tex]
According to stoichiometry:
3 mole of [tex]AgNO_3[/tex] react with 1 mole of [tex]AlCl_3[/tex]
0.025 moles of [tex]AgNO_3[/tex] react with =[tex]\frac{1}{3}\times 0.025=8.3\times 10^{-3}moles[/tex] of [tex]AlCl_3[/tex]
Thus [tex]AgNO_3[/tex] is the limiting reagent as it limits the formation of product and [tex]AlCl_3[/tex] is the excess reagent.
3 moles of [tex]AgNO_3[/tex] produce = 3 moles of [tex]AgCl[/tex]
0.025 moles of [tex]AgNO_3[/tex] produce =[tex]\frac{3}{3}\times 0.025=0.025moles[/tex] of [tex]AgCl[/tex]
Mass of [tex]AgCl[/tex] produced=[tex]moles\times {\text {Molar mass}}=0.025\times 143=3.6g[/tex]
Thus 3.6 g of [tex]AgCl[/tex] is produced.
The mass in grams of AgCl produced when 4.22 g of AgNO₃ react with 7.73 g of AlCl₃ is 3.56 g
From the question, the equation of the reaction is given as
3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agcl (s)
This can be written properly as
3AgNO₃ (aq) + AlCl₃ (aq) → Al(NO₃)₃ (aq) + 3AgCl (s)
From the balanced chemical equation above, we observe that
3 moles of AgNO₃ will react with 1 mole of AlCl₃ to yield 1 mole of Al(NO₃)₃ and 3 moles of AgCl.
To determine the mass of AgCl that would be produced when 4.22 g of AgNO₃ react with 7.73 g of AlCl₃,
First, we will convert the given masses to number of moles
Using the formula,
[tex]Number \ of \ moles = \frac{Mass}{Molar \ mass}[/tex]
For AgNO₃
Mass = 4.22 g
Molar mass = 169.87 g/mol
∴ [tex]Number \ of \ moles = \frac{4.22}{169.87}[/tex]
Number of moles of AgNO₃ present = 0.02484 moles
For AlCl₃
Mass = 7.73 g
Molar mass = 133.34 g/mol
∴ [tex]Number \ of \ moles = \frac{7.73}{133.34 }[/tex]
Number of moles of AlCl₃ present = 0.05797 moles
Since,
3 moles of AgNO₃ react with 1 mole of AlCl₃
Then,
0.02484 moles of AgNO₃ will react with [tex]\frac{0.02484}{3}[/tex] moles of AlCl₃ completely.
(NOTE: AgNO₃ is the limiting reagent and AlCl₃ is the excess reagent)
[tex]\frac{0.02484}{3} = 0.00828[/tex]
∴ Only 0.00828 moles of AlCl₃ will react.
Now, since
3 moles of AgNO₃ will react with 1 mole of AlCl₃ to yield 1 mole of Al(NO₃)₃ and 3 moles of 3AgCl
Therefore,
0.02484 moles of AgNO₃ will react with 0.00828 mole of AlCl₃ to yield 0.00828 mole of Al(NO₃)₃ and 0.02484 moles of AgCl
∴ 0.02484 moles of AgCl is produced during the reaction.
Now, we will convert this amount to grams
From the formula
[tex]Mass = Number \ of \ moles \times Molar \ mass[/tex]
Number of moles = 0.02484 moles
Molar mass = 143.32 g/mol
∴ Mass of AgCl produced = 0.02484 × 143.32
Mass of AgCl produced = 3.56 g
Hence, the mass in grams of AgCl produced when 4.22 g of AgNO₃ react with 7.73 g of AlCl₃ is 3.56 g
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