Question

Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volume of nitrogen dioxide is formed at 724 torr and 28.2° C by reacting 4.84 cm3 of copper (d = 8.95 g/cm3) with 227 mL of nitric acid (d = 1.42 g/cm3, 68.0% HNO3 by mass)?

Answer 1

Answer:

35.41 L

Explanation:

Given, Volume of Copper = 4.84 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 8.95 g/cm³ * 4.84 cm³ = 43.318 g

Mass of copper = 43.318 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{43.318\ g}{63.546\ g/mol}$

Moles of copper = 0.6817 moles

Given, Volume of nitric acid solution = 227 mL = 227 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 1.42 g/cm³ * 227 cm³ = 322.34 g

Also, Nitric acid is 68.0 % by mass. So,

Mass of nitric acid = $\frac {68}{100}\times 322.34\ g$ = 219.1912 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{219.1912\ g}{63.01\ g/mol}$

Moles of nitric acid = 3.4786 moles

According to the reaction,

$Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}$

1 mole of copper react with 4 moles of nitric acid

Thus,

0.6817 moles of copper react with 4*0.6817 moles of nitric acid

Moles of nitric acid required = 2.7268 moles

Available moles of nitric acid = 3.4786 moles

Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.6817 mole of copper on reaction forms 2*0.6817 moles of nitrogen dioxide

Moles of nitrogen dioxide = 1.3634 moles

Given:  

Pressure = 724 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,  

Pressure = 724 / 760 atm = 0.9526 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.2 + 273.15) K = 301.35 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9526 atm × V = 1.3634 mol × 0.0821 L.atm/K.mol × 301.35 K  

⇒V = 35.41 L