t 2.43 atm and 712 K, what volume of NO₂ gas forms from 47.98 g of 0₂? 4 NH3(g) + 70₂(g) 4 NO₂(g) + 6H₂O(1) irst, calculate the number of moles of NO₂ formed in this reaction and show the conversions required to solve this proble 7.98 g 0₂ x = mc Answer Bank 1 mol NO₂, 7 mol O₂, 4 mol NO₂, 1 mol O.

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At 2.43 atm and 712 K, what volume of NO₂ gas forms from 47.98 g of O₂?
4 NH3(g) + 70₂(g)
4 NO₂(g) + 6H₂O(1)
First, calculate the number of moles of NO₂ formed in this reaction and show the conversions required to solve this problem.
mol NO₂
=
47.98 g 0₂ X
Answer Bank
1 mol NO₂
4 mol NO₂
1 mol O₂
7 mol O₂
1 mol H₂O
18.02 g H₂O
1 mol NH3
4 mol NH3
17.03 g NH3
32.00 g 0₂
46.01 g NO₂
6 mol H₂O
Then, use the ideal gas law to calculate the volume of NO₂ gas formed.
2
Transcribed Image Text:At 2.43 atm and 712 K, what volume of NO₂ gas forms from 47.98 g of O₂? 4 NH3(g) + 70₂(g) 4 NO₂(g) + 6H₂O(1) First, calculate the number of moles of NO₂ formed in this reaction and show the conversions required to solve this problem. mol NO₂ = 47.98 g 0₂ X Answer Bank 1 mol NO₂ 4 mol NO₂ 1 mol O₂ 7 mol O₂ 1 mol H₂O 18.02 g H₂O 1 mol NH3 4 mol NH3 17.03 g NH3 32.00 g 0₂ 46.01 g NO₂ 6 mol H₂O Then, use the ideal gas law to calculate the volume of NO₂ gas formed. 2
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