How many grams of gas must be released from a 32.0 L sample of CO2(g) at STP to reduce the volume to 16.6 L at STP?
1 answer:
Answer:
30.3 g
Explanation:
At STP, 1 mol of any gas will occupy 22.4 L.
With the information above in mind, we <u>calculate how many moles are there in 32.0 L</u>:
- 32.0 L ÷ 22.4 L/mol = 1.43 mol
Then we <u>calculate how many moles would there be in 16.6 L</u>:
- 16.6 L ÷ 22.4 L/mol = 0.741 mol
The <u>difference in moles is</u>:
- 1.43 mol - 0.741 mol = 0.689 mol
Finally we <u>convert 0.689 moles of CO₂ into grams</u>, using its <em>molar mass</em>:
- 0.689 mol * 44 g/mol = 30.3 g
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