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geniusboy [140]

2 years ago

12

let 4 moles of methanol (liquid) combust in 3 moles of gaseous oxygen to form gaseous carbon dioxide and water vapor. Suppose th

is occurs in a chamber of fixed volume and fixed temperature. If the original pressure is 1.0 atm, what is the final pressure in the chamber. Express your answer in atm. Enter a numerical value, do not enter units. Assume liquids take up negligible volume.

1 answer:

kari74 [83]2 years ago

8 0

Answer:

The final pressure is 2.0 atm

Explanation:

<u>Step 1:</u> Data given

Number of moles methanol = 4 mol

Number of moles oxygen = 3 mol

original pressure is 1.0 atm

<u>Step 2:</u> The balanced equation

2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l)

<u>Step 3</u>: Calculate final pressure

For 2 moles of methanol consumed, we need 3 moles O2 to produce 2 moles CO2 and 4 moles H2O

We started with 3 moles of O2 gas.

Since methanol is not a gas, it doesn't count for the pressure.

V and T are fixed

This means the final pressure can be given by:

P2/P1 = n2/n1

with n2 = number moles of products

with n1 = number of moles of reactants

P2 = (6.0 mol*1.0 atm) / (3.0 mol )

P2 = Final pressure = 2.0 atm

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Calculate the number of kilojoules of energy required to convert 50.0 grams of solid DMSO initially at a temperature of 19.0°C t

GuDViN [60]

Answer:

20.79 kilojoules

Explanation:

Using Q = m×c×∆T

Where;

Q = Quantity of heat (J)

c = specific heat capacity of solid DMSO (1.80 J/g°C)

m = mass of DMSO

∆T = change in temperature

According to the provided information, m= 50g, initial temperature = 19.0°C, final temperature= 250.0°C

Q = m×c×∆T

Q = 50 × 1.80 × (250°C - 19°C)

Q = 90 × 231

Q = 20790 Joules

To convert Joules to kilojoules, we divide by 1000 i.e.

20790/1000

= 20.79 kilojoules

Hence, 20.79 kilojoules of energy is required to convert 50.0 grams of solid DMSO to gas.

4 0

2 years ago

A 25 gram(m) metal ball is heated to 200C(delta T) with 2330 Joules(q) of energy. What is the specific heat of the metal?

Dominik [7]

Answer:

The specific heat of the metal is 0.466 $\frac{J}{g*C}$

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Q= 2330 J
c= ?
m= 25 g
ΔT= 200 °C

Replacing:

2330 J= c*25 g* 200 °C

Solving:

$c=\frac{2330 J}{25 g* 200 C}$

c=0.466 $\frac{J}{g*C}$

<u><em>The specific heat of the metal is 0.466 </em></u> $\frac{J}{g*C}$ <u><em></em></u>

6 0

2 years ago

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A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

$\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}$

$P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}$

$P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times 35.0} = 0.776969038566$

∴ The final pressure P₂ ≈ 0.78 atm.

4 0

2 years ago

A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimete

N76 [4]

Answer:

The mass of water = 219.1 grams

Explanation:

Step 1: Data given

Mass of aluminium = 32.5 grams

specific heat capacity aluminium = 0.921 J/g°C

Temperature = 82.4 °C

Temperature of water = 22.3 °C

The final temperature = 24.2 °C

Step 2: Calculate the mass of water

Heat lost = heat gained

Qlost = -Qgained

Qaluminium = -Qwater

Q = m*c*ΔT

m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)

⇒with m(aluminium) = the mass of aluminium = 32.5 grams

⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C

⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C

⇒with m(water) = the mass of water = TO BE DETERMINED

⇒with c(water) = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C

32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9

-1742.1 = -7.95m

m = 219.1 grams

The mass of water = 219.1 grams

8 0

2 years ago

Using the equation, C5H12 + 8O2 Imported Asset 5CO2 + 6H2O, if an excess of pentane (C5H12) were supplied, but only 4 moles of o

ser-zykov [4K]

Answer: 3 <span>moles of water would be produced in present case.
</span>
Reason:
Reaction involved in present case is:
<span> C5H12 + 8O2 </span>→<span> 5CO2 + 6H2O

In above reaction, 1 mole of C5H12 reacts with 8 moles of oxygen to give 6 moles of water.

Thus, 4 moles of oxygen will react with 0.5 mole of C5H12, to generate 3 moles of H2O.</span>

7 0

2 years ago

Read 2 more answers