Ammonia, NH3, may react with oxygen to form nitrogen gas and water. 4NH3(aq)+302(g)→2N2(g)+6H2O(1) If 3.45 g NH3 reacts with 5.18g 02 and produces 0.950L N2 at 295 K and 1.00 bar, how many moles of N2N2 can be produced by each reactant? It may be useful to consult the periodic table. NH3: mol N2 02: mol N2 Based on the theoretical yields, which reactant is limiting? NH3(aq) 02(g) What is the percent yield of the reaction? percent yield:

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Ammonia, NH3, may react with oxygen to form nitrogen gas and water.
4NH3(aq)+302(g)→2N2(g)+6H2O(1)
If 3.45 g NH3 reacts with 5.18g 02 and produces 0.950L N2 at 295 K and 1.00 bar, how many moles of N2N2 can
be produced by each reactant? It may be useful to consult the periodic table.
mol N2
NH3:
02:
mol N2
Based on the theoretical yields, which reactant is limiting?
NH3(aq)
O2(g)
What is the percent yield of the reaction?
percent yield:
Transcribed Image Text:Ammonia, NH3, may react with oxygen to form nitrogen gas and water. 4NH3(aq)+302(g)→2N2(g)+6H2O(1) If 3.45 g NH3 reacts with 5.18g 02 and produces 0.950L N2 at 295 K and 1.00 bar, how many moles of N2N2 can be produced by each reactant? It may be useful to consult the periodic table. mol N2 NH3: 02: mol N2 Based on the theoretical yields, which reactant is limiting? NH3(aq) O2(g) What is the percent yield of the reaction? percent yield:
Expert Solution
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Recall the given reaction,                      4 NH3 aq  +  3 O2 g    2 N2 g  +  6 H2O l3.45 g NH3 reacts with 5.18 g O2 to produce 0.950 L N2 at 295 K and 1.00 bar .We have to determine - Moles of N2 produced by each reactant .                                            Limiting reactant .                                            Percent yield .

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hi, sorry the typing is a little cramed together can you seperate the answers 

NH3      molN2

O2          molN2         

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