Answer:
1. 9.45 g
2. 66.9%
Explanation:
The balanced equation for the reaction is given below:
Cu(s) + 4HNO3(aq) —> Cu(NO3)2(aq) + 2NO2(g) + 2 H2O(l)
Next, we shall determine the mass of Cu that reacted and the mass of Cu(NO3)2 produced from the balanced equation. This can be obtained as follow:
Molar mass of Cu = 63.5 g/mol
Mass of Cu from the balanced equation
= 1 × 63.5 = 63.5 g
Molar mass of Cu(NO3)2 = 63.5 + 2[14 + (3×16)]
= 63.5 + 2[14 + 48]
= 63.5 + 2[62]
= 63.5 + 124
= 187.5 g/mol
Mass of Cu(NO3)2 from the balanced equation = 1 × 187.5 = 187.5 g
Summary:
From the balanced equation above,
63.5 g of Cu reacted to produce 187.5 g of Cu(NO3)2.
1. Determination of the theoretical yield of Cu(NO3)2. This can be obtained as follow:
From the balanced equation above,
63.5 g of Cu reacted to produce 187.5 g of Cu(NO3)2.
Therefore, 3.20 g of Cu will react to produce = (3.20 × 187.5) / 63.5 = 9.45 g of Cu(NO3)2.
Thus, the theoretical yield of Cu(NO3)2 is 9.45 g.
2. Determination of the percentage yield of Cu(NO3)2.
Actual yield of Cu(NO3)2 = 6.32 g
Theoretical yield of Cu(NO3)2 = 9.45 g.
Percentage of the Cu(NO3)2 =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield of Cu(NO3)2 = 6.32/9.45 × 100
Percentage yield of Cu(NO3)2 = 66.9%
The theoretical yield of Cu(NO₃)₂ is 9.445 g
The percent yield of Cu(NO₃)₂ is 66.91%
From the question.
We are to determine the theoretical yield of Cu(NO₃)₂,
First and foremost, we will write the balanced chemical equation for the reaction properly
The balanced chemical equation for the reaction is
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)
This means
1 mole of Cu will react with 4 moles of HNO₃ to produce 1 mole of Cu(NO₃)₂, 2 moles of NO₂ and 2 moles of H₂O
To determine the theoretical yield of Cu(NO₃)₂,
First, we will determine the number of moles of Cu that reacted
Mass of Cu = 3.20 g
Molar mass of Cu = 63.546 g/mol
From the formula
[tex]Number\ of\ moles =\frac{Mass}{Molar\ mass}[/tex]
∴ Number of moles of Cu = [tex]\frac{3.20}{63.546}[/tex]
Number of moles of Cu present = 0.050357 moles
The number of moles of Cu that reacted is 0.050357 moles
Since,
1 mole of Cu will react with 4 moles of HNO₃ to produce 1 mole of Cu(NO₃)₂
Then,
0.050357 moles of Cu will react with 0.201428 moles of HNO₃ to produce 0.050357 moles of Cu(NO₃)₂
∴ 0.050357 moles of Cu(NO₃)₂ will be produced during reaction
Now, for the theoretical yield of Cu(NO₃)₂
From the formula
Mass = Number of moles × Molar mass
Molar mass of Cu(NO₃)₂ = 187.56 g/mol
∴ Theoretical yield of Cu(NO₃)₂ = 0.050357 × 187.56
Theoretical yield of Cu(NO₃)₂ = 9.445 g
Hence, the theoretical yield of Cu(NO₃)₂ is 9.445 g
For the percent yield of Cu(NO₃)₂
[tex]Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield } \times 100 \%[/tex]
From the question
Actual yield = 6.32 g
But,
Theoretical yield = 9.445 g
∴ Percent yield of Cu(NO₃)₂ = [tex]\frac{6.32}{9.445} \times 100\%[/tex]
Percent yield of Cu(NO₃)₂ = [tex]\frac{632}{9.445}\%[/tex]
Percent yield of Cu(NO₃)₂ = 66.91%
Hence, the percent yield of Cu(NO₃)₂ is 66.91%
Learn more here: https://brainly.com/question/13902065
