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2 years ago

The combustion of methyl alcohol in an abundant excess of oxygen follows the equation:

Chemistry

1 answer:

Anastaziya [24]2 years ago

6 0

Answer:

The % yield of CO2 is 69.1 %

Explanation:

Step 1: Data given

Mass of CH3OH = 3.85 grams

Mass of O2 = 6.15 grams

Mass of CO2 obtained = 3.65 grams

Molar mass of CH3OH = 32.04 g/mol

Molar mass O2 = 32.00 g/mol

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

2CH3OH + 3O2→ 2CO2 + 4H2O

Step 3: Calculate moles CH3OH

Moles CH3OH = mass CH3OH / molar mass CH3OH

Moles CH3OH = 3.85 grams / 32.04 g/mol

Moles CH3OH = 0.120 moles

Step 4: Calculate moles O2

Moles O2 = 6.15 grams / 32.00 g/mol

Moles O2 = 0.192 moles

Step 5: Calculate limtiting reactant

CH3OH is the limiting reactant. It will completely be consumed ( 0.120 moles). O2 is in excess. There will react 3/2 * 0.120 = 0.180 moles

There will remain 0.192 - 0.180 = 0.012 moles

Step 6: Calculate moles CO2

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 moles H2O

For 0.120 moles CH3OH we'll have 0.120 moles CO2

Step 7: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.120 moles * 44.01 g/mol

Mass CO2 = 5.28 grams

Step 8: Calculate % yield

% yield = (actual mass / theoretical mass) *100%

%yield = ( 3.65 grams / 5.28 grams ) *100%

% yield = 69.1 %

The % yield of CO2 is 69.1 %

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http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

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2 years ago

A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete

lys-0071 [83]

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j

8 0

2 years ago

A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of

allochka39001 [22]

Answer:

The answer to your question is C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

44 g of CO₂ --------------- 12 g of C

52.8 g --------------- x

x = (52.8 x 12)/44

x = 633.6/44

x = 14.4 g of C

12 g of C ------------------ 1 mol

14.4 g of C --------------- x

x = (14.4 x 1)/(12)

x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

18 g of H₂O --------------- 2 g of H

21.6 g of H₂O ------------- x

x = (21.6 x 2) / 18

x = 2.4 g of H

1 g of H -------------------- 1 mol of H

2.4 g of H ----------------- x

x = (2.4 x 1)/1

x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

= 6.4 g

16 g of O ---------------- 1 mol

6.4 g of O -------------- x

x = (6.4 x 1)/16

x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

C₃H₆O

8 0

2 years ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

7 0

2 years ago

Read 2 more answers

At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a

djverab [1.8K]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

$Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2$

Where P is the pressure of each compound in equilibrium.

If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:

H2S = 3.00 atm - 2X

H2 = 2X

S2: = X

Replacing:

$3.2x10^{-7} = (2X)^2X / (3-2X)^2$

$3.2x10^{-7} = 4X^3 / 9- 6X+4X^2$

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Solving for X:

X = 0.008945 atm

As in equilibrium, pressure of S2 is X, pressure is 0.008945 atm

7 0

2 years ago