Asked by katydid
How many grams of oxygen gas can be produced when 4.50 L of NO2 react according to the equation below at 427 ¨¬C and 1.15 atm?
4NO2(g) + 6H2O(g) ¡æ 7 O2(g) + 4NH3(g)
4NO2(g) + 6H2O(g) ¡æ 7 O2(g) + 4NH3(g)
Answers
Answered by
bobpursley
first figure the moles of NO2 (use the ideal gas law).
then you get 7/4 moles O2 for each mole of NO2. Change that to grams O2
then you get 7/4 moles O2 for each mole of NO2. Change that to grams O2
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