Question

calculate the volume occupied by 6.4g of oxygen, 02 at stp?,​

Answer 1

Answer:

4.9 L O₂

General Formulas and Concepts:

Atomic Structure

• Reading a Periodic Table
• Moles
• STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

• Using Dimensional Analysis

Explanation:

Step 1: Define

Identify variables

[Given] 6.4 g O₂ at STP

[Solve] L O₂

Step 2: Identify Conversions

[STP] 1 mol = 22.4 L

[PT] Molar Mass of O: 16.00 g/mol

Molar Mass of O₂: 2(16.00) = 32.00 g/mol

Step 3: Convert

1. [DA] Set up:                                                                                                       $\displaystyle 6.4 \ g \ O_2(\frac{1 \ mol \ O_2}{32.00 \ g \ O_2})(\frac{22.4 \ L \ O_2}{1 \ mol \ O_2})$
2. [DA] Divide/Multiply [Cancel out units]:                                                           $\displaystyle 4.48 \ L \ O_2$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

4.48 L O₂ ≈ 4.9 L O₂