Answer:
D. 44.2 g O₂
General Formulas and Concepts:
Gas Laws
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Stoichiometry
- Dimensional Analysis
- Mole Ratio
Explanation:
Step 1: Define
Identify given.
61.9 L O₂ at STP
Step 2: Convert
We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many moles of O₂ is present:
[tex]\displaystyle 61.9 \ \text{L} \ \text{O}_2 \bigg( \frac{1 \ \text{mol} \ \text{O}_2}{22.4 \ \text{L} \ \text{O}_2} \bigg) = 2.76339 \ \text{mol} \ \text{O}_2[/tex]
Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from moles to grams:
[tex]\displaystyle 2.76339 \ \text{mol} \ \text{O}_2 \bigg( \frac{16.00 \ \text{g} \ \text{O}_2}{1 \ \text{mol} \ \text{O}_2} \bigg) = 44.2143 \ \text{g} \ \text{O}_2[/tex]
Now we deal with sig figs. From the original problem, we are given 3 significant figures. Round your answer to the exact same number of sig figs:
[tex]\displaystyle 44.2143 \ \text{g} \ \text{O} \approx \boxed{ 44.2 \ \text{g} \ \text{O}_2 }[/tex]
∴ our answer is letter choice D.
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Topic: AP Chemistry
Unit: Stoichiometry
