Suppose 3.20 g of copper are reacted with excess nitric acid according to the given equation, and 6.32 g Cu(NO3)2 product are obtained. Cu(s) + 4 HNO3 (aq) --> Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O(l)

What is the theoretical yield of Cu(NO3)2? In g
What is the percent yield of Cu(NO3)2? In %

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

∴ Number of moles of Cu =

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

From the question

Actual yield = 6.32 g

But,

Theoretical yield = 9.445 g

∴ Percent yield of Cu(NO₃)₂ =

Percent yield of Cu(NO₃)₂ =

Percent yield of Cu(NO₃)₂ = 66.91%

Hence, the percent yield of Cu(NO₃)₂ is 66.91%

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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%

fe 55.85 g/mol and h2o 18.02 g/mol.

16

explanation:

the marks go by 4s