Question

A reaction between liquid reactants takes place at 30.0 °C in a sealed, evacuated vessel with a measured volume of 20.0 L. Measurements show that the reaction produced 17. g of carbon monoxide gas. Calculate the pressure of carbon monoxide gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.

Answer 1

Answer:

The correct answer is 0.75 atm

Explanation:

We have carbon monoxide gas (CO) at the following conditions:

T= 30ºC = 303 K

V= 20.0 L

m = 17 g

The molecular weight of CO (MM CO) is the following:

MM CO= molar mass of C + molar mass of O = 12 g/mol + 16 g/mol = 28 g/mol

We calculate the number of moles (n) as follows:

n= m/MM CO = 17 g/28 g/mol = 0.61 mol

Finally we use the ideal gases equation to calculate the pressure (P):

P x V = n x R x T

P = (n x R x T)/V

P= (0.61 mol x 0.082 L.atm/K.mol x 303 K)/20.0 L

P= 0.75 atm

Answer 2

Answer:

The pressure of the carbon monoxide gas is 0.75 atm

Explanation:

Step 1: Data given

Temperature = 30.0 °C = 303 K

Volume = 20.0 L

MAss of carbon monoxide gas = 17.0 grams

Molar mass of CO = 28.01 g/mol

Step 2: Calculate moles CO

Moles CO = mass CO / molar mass CO

Moles CO = 17.0 grams / 28.01 g/mol

Moles CO = 0.607 moles

Step 3: Calculate the pressure of the carbon monoxide gas

p*V = n*R*T

⇒with p = the pressure of the carbon monoxide gas = TO BE DETERMINED

⇒with v = the volume of the vessel = 20.0 L

⇒with n = the number of moles of CO = 0.607 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 303 K

p = (n*R*T) / V

p = (0.607 moles * 0.08206 L*atm/mol*K * 303 K) / 20.0 L

p = 0.754 atm ≈ 0.75 atm

The pressure of the carbon monoxide gas is 0.75 atm