Question

sometimes in lab we collect the gas formed by a chemical reaction over water (see sketch at right). this makes it easy to isolate and measure the amount of gas produced. suppose the o₂ gas evolved by a certain chemical reaction taking place at 35.0°c is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 59.8 ml. calculate the mass of o₂ that is in the collection tube. round your answer to 2 significant digits. you can make any normal and reasonable assumption about the reaction conditions and the nature of the gases.

Explanation:

Step1: Find the pressure of \(O_2\) gas

Assume the total pressure \(P_{total}= 1\space atm\). The vapor - pressure of water at \(T = 35.0^{\circ}C\) is \(P_{H_2O}=42.2\space mmHg\). Convert \(P_{total}\) to \(mmHg\): \(P_{total}=760\space mmHg\).
Using Dalton's law of partial pressures \(P_{O_2}=P_{total}-P_{H_2O}\)
\(P_{O_2}=760 - 42.2=717.8\space mmHg\). Convert to \(atm\): \(P_{O_2}=\frac{717.8}{760}\space atm\approx0.9445\space atm\)

Step2: Convert volume and temperature

The volume \(V = 59.8\space mL=0.0598\space L\). The temperature \(T=(35.0 + 273.15)K=308.15\space K\)

Step3: Use the ideal gas law \(PV=nRT\) to find the number of moles of \(O_2\)

The ideal gas constant \(R = 0.0821\space L\cdot atm\cdot K^{-1}\cdot mol^{-1}\). From \(PV=nRT\), we can solve for \(n\): \(n=\frac{P_{O_2}V}{RT}\)
Substitute \(P_{O_2}=0.9445\space atm\), \(V = 0.0598\space L\), \(R = 0.0821\space L\cdot atm\cdot K^{-1}\cdot mol^{-1}\), and \(T = 308.15\space K\)
\(n=\frac{0.9445\times0.0598}{0.0821\times308.15}\)

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Step4: Calculate the mass of \(O_2\)

The molar mass of \(O_2\) is \(M = 32.0\space g/mol\). Using \(m=nM\), substitute \(n = 0.00223\space mol\) and \(M = 32.0\space g/mol\)
\(m=0.00223\times32.0 = 0.0714\space g\approx0.071\space g\)

Answer:

\(0.071\space g\)