Question

what is the mass of 61.9 l of oxygen gas collected at stp?a 88.4 g o₂b 44.2 g o₂c 2.76 g o₂d 122 g o₂

Explanation:

Step1: Recall STP molar volume

At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L. So, first, find moles of \(O_2\) using volume.
Moles of \(O_2\), \(n = \frac{V}{22.4\ L/mol}\) where \(V = 61.9\ L\).
\(n=\frac{61.9\ L}{22.4\ L/mol}\approx2.763\ mol\)

Step2: Calculate molar mass of \(O_2\)

Molar mass of \(O\) is 16.00 g/mol, so for \(O_2\), molar mass \(M = 2\times16.00\ g/mol = 32.00\ g/mol\).

Step3: Find mass using \(m = n\times M\)

\(m = 2.763\ mol\times32.00\ g/mol\approx88.4\ g\)

Answer:

a. \(88.4\ g\ O_2\)