Question

If the partial pressure of a gas produced in a reaction was determined to be 0.35 atm in a 161.0 ml container and at a temperature of 25.0°c, how many moles of gas were produced?

Answer 1

Thank you for posting your question here at brainly. I hope the answer will help you. The number of moles of gas were produced the answer will be 0.0023 mol. 

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Answer 2

Answer: The number of moles of gas produced are 0.0023 moles.

Explanation:

To calculate the number of moles of gas, we use the equation given by ideal gas, which follows:

$PV=nRT$

where,

P = pressure of the gas = 0.35 atm

V = Volume of the gas = 161.0 mL = 0.161 L    (Conversion factor: 1 L = 1000 mL)

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of gas = ?

Putting values in above equation, we get:

$0.35atm\times 0.161L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{0.35\times 0.161}{0.0821\times 298}=0.0023mol$

Hence, the number of moles of gas produced are 0.0023 moles.