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 25°c and a pressure of 741 mmhg. if the wet o₂ gas formed occupies a volume of 7.29 l the number of grams of o₂ formed is
g. the vapor pressure of water is 23.8 mmhg at 25°c.
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Explanation:

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

According to Dalton's law of partial pressures, \(P_{total}=P_{O_2}+P_{H_2O}\). So, \(P_{O_2}=P_{total}-P_{H_2O}\).
Given \(P_{total} = 741\space mmHg\) and \(P_{H_2O}=23.8\space mmHg\) at \(25^{\circ}C\), then \(P_{O_2}=(741 - 23.8)\space mmHg=717.2\space mmHg\).
Convert pressure to atm: \(P=\frac{717.2}{760}\space atm\approx0.9437\space atm\) (since \(1\space atm = 760\space mmHg\)).

Step2: Use the ideal gas law \(PV = nRT\) to find the number of moles of \(O_2\)

The ideal gas law is \(n=\frac{PV}{RT}\).
Given \(V = 7.29\space L\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), and \(T=(25 + 273)K=298K\).
Substitute the values: \(n=\frac{0.9437\times7.29}{0.0821\times298}\).
First, calculate the numerator: \(0.9437\times7.29 = 6.88\).
Then, calculate the denominator: \(0.0821\times298=24.47\).
So, \(n=\frac{6.88}{24.47}\approx0.281\space mol\).

Step3: Calculate the mass of \(O_2\)

The molar mass of \(O_2\) is \(M = 32\space g/mol\).
Using the formula \(m=n\times M\), substitute \(n = 0.281\space mol\) and \(M = 32\space g/mol\).
\(m=0.281\times32\space g = 8.99\space g\approx9.0\space g\).

Answer:

\(9.0\)