Ammonia, nh3nh3 , can react with oxygen to form nitrogen gas and water. 4nh3(aq)+3o2(g)⟶2n2(g)+6h2o(l) 4nh3(aq)+3o2(g)⟶2n2(g)+6h2o(l) if 2.35 g2.35 g nh3nh3 reacts with 3.53 g3.53 g o2o2 and produces 0.650 l0.650 l n2n2 , at 295 k295 k and 1.01 bar1.01 bar , which reactant is limiting? o2(g)o2(g) nh3(aq)nh3(aq) calculate the percent yield of the reaction. percent yield:

36.37% is the percent yield of the reaction.

Explanation:

1)0.650 L nitrogen gas  , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P=  1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T =  295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

2) Moles of ammonia gas=

Moles of oxygen gas =

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

Percentage yield of the reaction:

36.37% is the percent yield of the reaction.

idk what the question is about

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )