Answer
B) 47.0
Step-by-step explanation
If at STP, exactly 1 mole of any gas occupies 22.4 L
Then, 2.1 moles of O₂ gas at STP will occupy x L container
To get x L, cross multiply and divide both sides by 1 mole.
![x\text{ }L\text{ }container=\frac{2.1\text{ }mol}{1\text{ }mol}*22.4\text{ }L=47.0\text{ }L](https://img.qammunity.org/2023/formulas/chemistry/college/cazkwgp46tbwhv1s0j83pt6l3pm4ht63iq.png)
Therefore, the size of the container (in Liters) needed to hold 2.1 mol O₂ gas at STP is 47.0 L