Answer to Question #164327 in General Chemistry for muna mohamed

Question

1. At what temperature will 0.654 moles of neon gas occupy 12.30 L at 1.95 atm?
2. A sample of argon gas at STP occupies 56.2 L. Determine the number of moles of argon.
3. A gas is compressed in a 25 L storage tank under a pressure of 150 kPa and a temp of
4. 21°C. How many moles of gas are contained in the tank?
5. What volume will 45 grams of hydrogen gas occupy at 1.05 atm and 25°C?
6. At what temperature will 5.00 g of Cl2 exert a pressure of 900 mm Hg at a volume of 750.0 mL?
7. Ammonium nitrate decomposes explosively upon heating according to the following equation:
8.             NH4NO3(s)  N2  (g) + O2 (g) + H2O (g) 
9. Calculate the total volume of gas produced at 125°C and 748 mm Hg when 1.55 kg of ammonium nitrate completely decomposes. Balance the equation! 

Answer 1

Q164327

1) At what temperature will 0.654 moles of neon gas occupy 12.30 L at 1.95 atm?

Solution:

T = unknown, n = 0.654 mol, V = 12.30 L, P = 1.95 atm .

We can use the ideal gas equation, PV = nRT, and find the temperature of the gas.

$PV=nRT ;$ where R = 0.08206 L-atm/mol.K

substitute all the given information in the ideal gas equation we have

$1.95 atm.12.30 L = 0.654mol * 0.08206L.atm/mol.K*T$

$\frac{23.985 L.atm }{0.05367L.atm/K}= \frac{0.05367L.atm/K*T }{0.05367L.atm/K}$

divide both the side by 0.05367 L.atm/K we have

T = 446.9K. = 446.9K - 273.15 = 173.8 °C

Hence the neon gas will be at 447 Kelvin temperature.

which in degree celsius will be 174 °C

2) A sample of argon gas at STP occupies 56.2 L. Determine the number of moles of argon.

Solution:

At STP, T = 273.15K and P = 1atm.

We are given V = 56.2 L.

Gas constant, R = 0.08206 L.atm/mol.K.

Substitute all this information in the ideal gas equation, PV = nRT we have

$$PV = nRT$$

$1atm * 56.2L = n * 0.08206 L.atm/mol.K* 273.15K.$

$56.2L.atm = n * 22.415 L.atm/mol$

divide both the side by 22.415L.atm/mol, we have

$\frac{56.2L.atm}{ 22.415 L.atm/mol }=\frac{ n * 22.415 L.atm/mol }{22.415 L.atm/mol}$

n = 2.5073 moles of Argon.

in the question, we are given 56.2 L in 3 significant figure, so our final answer must also be in

3 significant figure.

Hence there would be 2.51 moles of Argon present in the given sample.

3) A gas is compressed in a 25 L storage tank under a pressure of 150 kPa and a temp of

21°C. How many moles of gas are contained in the tank?

Solution:

We are given Volume, V = 25L, Pressure, P = 150 kPa

Temperature T = 21°C.

In the ideal gas equation, pressure must be used in the 'átm' unit and temperature must be in the Kelvin unit.

$Pressure \space P \space in\space atm = 150 kPa * \frac{1atm }{101.325 kPa } = 1.4804 atm.$

$Temperature\space in\space Kelvin = 21°C = 21 + 273.15 = 294.15K$

Gas constant, R = 0.08206 L.atm/mol.K

substitute all this information in the ideal gas equation we have

$$PV = nRT$$

$1.480atm * 25L = n * 0.08206 L.atm/mol.K * 294.15K$

$37L.atm = n * 22.415 L.atm/mol$

divide both the side by 22.415L.atm/mol, we have

$\frac{37L.atm}{ 22.415 L.atm/mol }=\frac{ n * 22.415 L.atm/mol }{22.415 L.atm/mol}$

$n = \frac{37L.atm}{ 22.415 L.atm/mol }$

n = 1.6507 moles.

in the question, we are given pressure and volume both in 2 significant figure, so our final answer must

also, be in 2 significant figures.

Hence there would be 1.7 moles of gas in the tank.

Question # 4 is not present

5) What volume will 45 grams of hydrogen gas occupy at 1.05 atm and 25°C?

Solution:

Here instead of 'moles of hydrogen gas' we are given the grams of hydrogen gas.

So we will first convert 45 grams of H₂ gas to moles by using the molar mass of H2 gas.

molar mass of H₂ = 2 * atomic mass of H = 2 * 1.00794 g/mol

= 2.016 g/mol

moles of H₂ $= 45 \space g \space of \space H_2 * \frac{1 \space mol \space H_2}{2.016 \space g of \space H_2}$ $= 22.32\space mol\space of \space H_2$

So now we have

n = 22.32 moles , P = 1.05 atm .

T = 25°C = 25 + 273.15 = 298.15K.

Gas constant, R = 0.08206 L.atm/mol.K

substitute this information in the ideal gas equation we have

$$PV = nRT$$

1.05 atm * V = 22.32 mol * 0.08206 L.atm/mol.K * 298.15K

1.05 atm * V = 546.09 L.atm.

divide both the side by 1.05 atm, we have

$\frac{1.05 atm * V }{1.05 atm }= \frac{546.09 L.atm. }{1.05 atm }$

V = 520.08 L.

In the given question the quantity with the least number of the significant figure is 45 grams.

It is in 2 significant figures. So our final answer must also be in 2 significant figures.

Hence the volume occupied by the hydrogen gas will be 520 L.

6) At what temperature will 5.00 g of Cl₂ exert a pressure of 900 mm Hg at a volume of 750.0 mL?

Solution.

Here too we are given a mass of Cl₂ gas. So first we will have to convert it to grams by using

the molar mass of Cl₂.

molar mass of Cl₂ = 2 * atomic mass of Cl = 2 * 35.453 g/mol

= 70.906 g/mol

moles of Cl₂ $= 5.00 g of Cl_2 * \frac{1 mol\space of\space Cl_2 }{70.906\space g\space\space of\space Cl_2 }$ $= 0.07052 \space moles\space of\space Cl_2$

Convert 750.0mL to 'L' and 900 mm Hg to 'atm' .

$volume\space in\space 'L' = 750.0mL * \frac{1L}{1000mL} = 0.750 L$

$pressure\space in\space 'atm' = 900 mm Hg * \frac{1atm }{760 mm Hg } = 1.184 atm.$ ( since, 1atm = 760 mm Hg)

So now we have

moles of Cl₂ = 0.07052 mol

volume occupied by Cl₂ = 0.750 L

pressure = 1.184 atm.

gas constant, R = 0.08206 L.atm/mol.K

plug all this information in the idea gas equation we have

$$PV = nRT$$

$1.184 atm * 0.750L = 0.07052 mol * 0.08206 L.atm/mol.K * T.$

0.8882 L.atm = 0.005787 L-atm/K * T

divide both the side by 0.005784 L.atm/mol, we have

$\frac{0.8882 L.atm}{0.005787 L-atm/K} = \frac{0.005787 L-atm/K * T }{0.005787 L-atm/K}$

$T =\frac{0.8882 L.atm}{0.005787 L-atm/K} = 153.48 K .$

which in 3 significant figure is 153 K or in

degree celsius is = 153.45 K - 273.15 = -119.66 °C = -120 °C

Hence the temperature of the Cl₂ gas will be 153 K or -120 °C.

7) Ammonium nitrate decomposes explosively upon heating according to the following equation:

            NH₄NO₃(s) ----->  N₂  (g) + O₂ (g) + H₂O (g) 

Calculate the total volume of gas produced at 125°C and 748 mm Hg when 1.55 kg of ammonium nitrate completely decomposes. Balance the equation! 

Solution:

Our first step will be to balance the given reaction.

 NH₄NO₃(s) ----->  N₂  (g) + O₂ (g) + H₂O (g) 

We will first balance the element which is not present in a single elemental form.

So N and O will be balanced afterward. First, we will balance hydrogen.

There is 4 H on the left side and 2 on the right side, so to balance the H write 2 in front of H₂O.

This will balance H on both the side.

NH₄NO₃(s) ----->  N₂  (g) + O₂ (g) + 2 H₂O (g) 

Nitrogen is already balanced here.

There are 3 O on the left side and 4 on the right side.

to balance the oxygen we will have to write a proper coefficient in front of O2.

placing a fraction 1/2 in front of O2 will make oxygen equal on both the side.

NH₄NO₃(s) -----> N₂  (g) + 1/2 O₂ (g) + 2 H₂O (g) 

We can't keep fractions in our balanced equation. To remove fraction multiply the whole reaction by 2.

2 * ( NH₄NO₃(s) -----> N₂  (g) + 1/2 O₂ (g) + 2 H₂O (g) )

2 NH₄NO₃(s) ----->  2 N₂  (g) + 1 O₂ (g) + 2 H₂O (g)

So now this reaction is balanced.

Step 1: Convert 1.55 kg of NH4 NO3 to moles by using the molar mass of NH₄NO₃

1.55 kg NH₄NO₃ in grams = 1.55 kg * 1000g/1kg = 1550 grams of NH₄NO₃

molar mass of NH₄ NO₃ = 2 * atomic mass of N + 3 * atomic mass of O + 4 * atomic mass of H

= 2 * 14.007 g/mol + 3 * 15.999 g/mol + 4 * 1.00794 g/mol

= 80.0426g/mol

moles of NH4NO3 = 1550 g of NH₄NO₃ * 1mol NH₄NO₃ / 80.0426 g NH₄NO₃

= 19.365 mol NH₄NO₃

in the balanced reaction, we can see that

2 NH₄NO₃(s) ----->  2 N₂  (g) + 1 O₂ (g) + 2 H₂O (g)

2 moles of NH4 NO3 gives total = 2 + 1 + 2 = 5 moles of gas.

so 19.365 moles of NH4NO3 will give

= 19.365 mol NH4NO3 * 5 mol gas / 2 mol NH4 NO3 = 48.412 mol gas

Step 2: To find the volume occupied by the gas.

We have moles of gas = 48.412 moles

T = 125°C = 398.15K

Pressure P = 748 mm Hg, which in átm = 748mmHg * 1atm / 760 mm Hg = 0.984 atm .

Volume = unknown

gas constant R = 0.08206 L.atm/mol.K

moles of gas = 48.412 mol

plug all this information in the ideal gas equation, PV = nRT

0.984 atm * V = 48.412 mol * 0.08206 L.atm/mol.K * 398.15K

0.984atm * V = 1581.7 L-atm

divide both the side by 0.984 atm we have

V = 1581.73L-atm/ 0.984 atm = 1607.4 L

Hence the volume occupied by the combusted gas after completely decompostion of NH4NO3 is

1600Liters.