Answer:
NH3 is the limiting reactant
The % yield is 36.1 %
Explanation:
<u>Step 1: </u>Data given
Mass of NH3 = 2.15 grams
Mass of O2 = 3.23 grams
Molar mass of NH3 = 17.03 g/mol
Molar mass of O2 = 32 g/mol
volume of N2 produced = 0.550 L
Temperature = 295 K
Pressure = 1.00 atm
<u>Step 2:</u> The balanced equation:
4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)
<u>Step 3:</u> Calculate moles of NH3
Moles NH3 = Mass NH3 / Molar Mass NH3
Moles NH3 = 2.15 grams / 17.03 g/mol
Moles NH3 = 0.126 moles
<u>Step 4:</u> Calculate moles of O2
Moles O2 = 3.23 grams / 32 g/mol
Moles O2 = 0.101 moles
<u>Step 5: </u>Calculate the limiting reactant
For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O
NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).
O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed
There will remain 0.101 - 0.945 = 0.0065 moles of O2
<u>Step 6:</u> Calculate moles of N2
For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O
For 4 moles NH3 , we'll have 2 moles of N2 produced
For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.
<u>Step 7</u>: Calculate volume of N2 produced
p*V = n*R*T
⇒ with p = the pressure of the gas = 1.00 atm
⇒ with V = the volume = TO BE DETERMINED
⇒ with n = the number of moles N2 = 0.063 moles
⇒ with R = the gasconstant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 295
V = (nRT)/p
V = (0.063*0.08206*295)/1
V = 1.525 L = theoretical yield
<u>Step 8:</u> Calculate the % yield
% yield = actual yield / theoretical yield
% yield = (0.550 L / 1.525 L)*100%
% yield = 36.1 %
