Question

When copper metal is added to nitric acid, the following reaction takes place
Cu (s) + 4 HNO₃ (aq) → Cu(NO₃)₂ (aq) + 2 H₂O (l) + 2 NO₂ (g)
Calculate the volume of NO₂ gas collected over water at 25.0 °C when 2.01 g of copper is added to excess nitric acid if the total pressure is 726 mm Hg. The vapor pressure of water at 25.0 °C is 23.8 mm Hg

Answer 1

The volume of NO₂ gas collected over water at 25.0 °C is 1.67 L

We'll begin by calculating the number of mole of Cu. This can be obtained as follow:

Mass of Cu = 2.01 g

Molar mass of Cu = 63.55 g/mol

Mole of Cu =?

Mole = mass / molar mass

Mole of Cu = 2.01 / 63.55

Mole of Cu = 0.0316 mole

Next, we shall determine the number of mole of NO₂ produced by 2.01 g (i.e 0.0316 mole) of Cu. This can be obtained as follow:

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2H₂O(l) + 2NO₂(g)

From the balanced equation above,

1 mole of Cu reacted to produce 2 moles of NO₂.

Therefore,

0.0316 mole of Cu will react to produce = 0.0316 × 2 = 0.0632 mole of NO₂.

Finally, we shall determine the volume of NO₂ gas obtained from the reaction. This can be obtained as follow:

Number of mole of NO₂ (n) = 0.0632 mole

Temperature (T) = 25 °C = 25 + 273 = 298 K

Total pressure = 726 mm Hg

Vapor pressure of water = 23.8 mm Hg

Pressure of NO₂ (P) = 726 – 23.8 = 702.2 / 760 = 0.924 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

PV = nRT

0.924 × V = 0.0632 × 0.0821 × 298

0.924 × V = 1.54623856

Divide both side by 0.924

$V = \frac{1.54623856}{0.924}$

V = 1.67 L

Therefore, the volume of NO₂ collected over water at 25.0 °C is 1.67 L

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Answer 2

The chemical reaction is defined as the reaction of reactants to form products. The reaction between copper and nitric acid will result in the formation of copper(II)nitrate, water, and nitrogen dioxide. The volume of NO$_2$ collected will be 1.67 L.

Give that,

• Mass of Cu = 2.01 g
• Molar mass of Cu = 63.55 g/mol

• $\begin{aligned}\text{mole}&=\dfrac{\text{mass}}{\text{molar mass}}\\\text{Mole of Cu}&=\dfrac{2.01}{ 63.55}\end$  
• Mole of Cu = 0.0316 mole

The number of moles produced by nitrogen dioxide (2.01 g) will be:

• $\text {Cu}_\text {(s)} + \text{4 HNO}_3_\text{(aq)}\rightarrow\text{Cu}\text{(NO}_3)_2_\text{(aq)} + \text{2 H}_2\text O_\text{(l)} + \text{2 NO}_2_\text{(g)}$

Now, from the equation,

• 0.316 moles of copper will produce = 2 x  0.0316 = 0.0632 mole of NO₂.

Now, from the ideal gas equation,

• PV = nRT
• 0.924 × V = 0.0632 × 0.0821 × 298
• 0.924 × V = 1.54623856
• Divide both side by 0.924

• $\text{V}&=\dfrac{1.546238}{0.924}\\\\\text V &= 1.67 L$

Therefore, the volume of nitrogen dioxide collected over water at 25-degree celcius is 1.67 L.

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