Question

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:

Answer 1

Answer:

36.37% is the percent yield of the reaction.

Explanation:

$4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)$

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:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol$

2) Moles of ammonia gas=$\frac{2.53 g}{17 g/mol}=0.1488 mol$

Moles of oxygen gas =$\frac{3.53 g}{32 g/mol}=0.1101 mol$

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

Then,0.1101 mol of oxygen will react with:

$\frac{4}{3}\times 0.1101 mol=0.1468 mol$ 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:

$\frac{2}{3}\times 0.1101 mol=0.0734 mol$ of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

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

Percentage yield:

$\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the reaction:

$\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%$

36.37% is the percent yield of the reaction.