Question

oxygen gas can be prepared by heating potassium chlorate according to the following equation:
2kclo₃(s) → 2kcl(s) + 3o₂(g)
the product gas, o₂, is collected over water at a temperature of 20°c and a pressure of 757.0 mmhg. if the wet o₂ gas formed occupies a volume of 5.37 l, the number of grams of o₂ formed is g. the vapor pressure of water is 17.5 mmhg at 20°c.

Explanation:

Step1: Calculate the pressure of dry \(O_2\)

The total pressure \(P_{total}=757.0\space mm\space Hg\), and the vapor pressure of water \(P_{H_2O}=17.5\space mm\space Hg\).
Using Dalton's law of partial pressures \(P_{O_2}=P_{total}-P_{H_2O}\)
\(P_{O_2}=757.0 - 17.5=739.5\space mm\space Hg\).
Convert pressure to atm: \(P=\frac{739.5}{760}\space atm\approx0.973\space atm\)
Volume \(V = 5.37\space L\), temperature \(T=(20 + 273.15)\space K=293.15\space K\)

Step2: Use the ideal gas law \(PV=nRT\)

\(R = 0.0821\space L\cdot atm/(mol\cdot K)\)
From \(PV=nRT\), we can solve for \(n\) (moles of \(O_2\)): \(n=\frac{PV}{RT}\)
Substitute the values: \(n=\frac{0.973\times5.37}{0.0821\times293.15}\)
\(n=\frac{5.235}{24.06}\approx0.218\space mol\)

Step3: Calculate the mass of \(O_2\)

The molar mass of \(O_2\) is \(M = 32.00\space g/mol\)
Using \(m=n\times M\), \(m=0.218\times32.00 = 6.98\space g\)

Answer:

\(6.98\space g\)