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1 year ago

What volume (in L) will a 32 g sample of butane gas, C4H10(g), occupy at a temperature of 45.0 oC and a pressure of 728 mm Hg?

Chemistry

1 answer:

larisa86 [58]1 year ago

6 0

Answer:

15.0 L

Explanation:

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (mmHg)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)

-----> T = temperature (K)

To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

32 grams C₄H₁₀ 1 moles
------------------------- x ----------------------- = 0.551 moles C₄H₁₀
58.124 grams

P = 728 mmHg R = 62.36 L*mmHg/mol*K

V = ? L T = 45.0 °C + 273.15 = 318.15 K

n = 0.551 moles

PV = nRT

(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)

(728 mmHg)V = 10922.7632

V = 15.0 L

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3 years ago

Rank the following compounds in order of decreasing boiling point: potassium fluoride (KF), acetylene (C2H2), and formaldehyde (

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Answer: $KF$ > $CH_2O$ > $C_2H_2$

Explanation:

The order of boiling point depends upon the type of interactions present between the molecules.

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2 years ago

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2 years ago

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3 years ago

15.00 grams of Chromium react with 15.00 grams of hydrobromic acid. Calculate the theoretical yield of the reaction. At STP what

s2008m [1.1K]

Answer:

(a) 18.03 g

(b) 2.105 L

Step-by-step explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.

M_r: 52.00 80.91 291.71

2Cr + 6HBr ⟶ 2CrBr₃ + 3H₂

Mass/g: 15.00 15.00

Step 2. Calculate the moles of each reactant

Moles of Cr = 15.00 × 1/52.00

Moles of Cr = 0.2885 mol Cr

Moles of HBr = 15.00 × 1/80.91

Moles of HBr = 0.1854 mol HBr ×

Step 3. Identify the limiting reactant

Calculate the moles of CrCl₃ we can obtain from each reactant.

From Cr:

The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr

Moles of CrBr₃ = 0.2885 × 2/2

Moles of CrBr₃ = 0.2885 mol CrCl₃

From HBr:

The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.

Moles of CrBr₃ = 0.1854 × 2/6

Moles of CrBr₃ = 0.061 80 mol CrBr₃

The limiting reactant is HBr because it gives the smaller amount of CrBr₃.

Step 4. Calculate the theoretical yields of CrBr₃ and H₂.

Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1

Theoretical yield of CrBr₃ = 18.03 g CrCl₃

The molar ratio is 3 mol H₂:6 mol HBr

Theoretical yield of H₂ = 0.1854 × 3/6

Theoretical yield of H₂ = 0.092 70 mol H₂

Step 5. Calculate the volume of H₂ at STP

STP is 1 bar and 0 °C.

The molar volume of a gas at STP is 22.71 L.

Volume = 0.092 70 × 22.71/1

Volume = 2.105 L

Step 6. Calculate the percent yield

% Yield = actual yield/theoretical yield × 100 %

Actual yield = 15.35 g

% yield = 15.35/18.03 × 100

% yield = 85.15 %

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3 years ago