Question

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a cylinder contains 22.5 l of oxygen gas at a pressure of 2.10 atm and a temperature of 305 k.
part a
what is the mass of oxygen in the cylinder?
express the mass in grams to three significant figures.
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Explanation:

Step1: Use ideal gas law to find moles

The ideal gas law is \(PV = nRT\). Rearranging for \(n\) (moles), we get \(n=\frac{PV}{RT}\).
Given \(P = 2.10\space atm\), \(V=22.5\space L\), \(T = 305\space K\), and \(R=0.0821\space L\cdot atm/(mol\cdot K)\).
Substitute values: \(n=\frac{2.10\times22.5}{0.0821\times305}\).
Calculate numerator: \(2.10\times22.5 = 47.25\).
Calculate denominator: \(0.0821\times305=25.0405\).
So \(n=\frac{47.25}{25.0405}\approx1.887\space mol\).

Step2: Convert moles to mass

Molar mass of \(O_2\) is \(M = 32.0\space g/mol\).
Use formula \(m=n\times M\).
Substitute \(n = 1.887\space mol\) and \(M = 32.0\space g/mol\).
\(m=1.887\times32.0\).
\(m = 60.4\space g\).

Answer:

\(60.4\space g\)