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olga nikolaevna [1]
3 years ago
9

What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of

an aqueous solution of nitric acid (d = 1.42 g/ cm3, 68.0% by mass) ? Cu(s) + 4 HNO3 (aq) → Cu(NO3)2(aq) + 2NO2(g) + 2 H2O(ℓ)
Chemistry
1 answer:
weqwewe [10]3 years ago
4 0

Answer:

25.76 L

Explanation:

Given, Volume of Copper = 3.56 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 * 3.56 cm³ = 31.862 g

Mass of copper = 31.862 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{31.862\ g}{63.546\ g/mol}

<u>Moles of copper = 0.5014 moles </u>

Given, Volume of nitric acid solution = 200 mL = 200 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 * 200 cm³ = 284 g

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

Mass of nitric acid = \frac {68}{100}\times 284\ g = 193.12 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{193.12\ g}{63.01\ g/mol}

<u>Moles of nitric acid = 3.0649 moles </u>

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,  

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

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

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,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

<u>Moles of nitrogen dioxide = 1.0028 moles </u>

Given:  

Pressure = 735 torr

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

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

So,  

Pressure = 735 / 760 atm = 0.9632 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:

0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K  

<u>⇒V = 25.76 L</u>

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