Answer:
There will be produced 5.398 grams of NH3
Explanation:
Step 1: Data given
Partial pressure H2 = 536 torr
Partial pressure of NO = 602 torr
Temperature = 150.0°C
Volume = 25.0 L
Reaction yield = 93.5 %
Step 2: The balanced equation
3H2 + 2NO → 2NH3 + O2
Step 3: Calculate moles of H2
p*V = n*R*T
⇒ with p = the pressure of H2 = 536 torr = 536/760 = 0.70526 atm
⇒ with V = the volume = 25.0 L
⇒ with n = the number of moles of H2 = TO BE DETERMINED
⇒ with R= the gas constant = 0.08206 L*atm/mol*K
⇒ with T =the temperature = 150.0 °C = 423 K
n= (0.70526*25.0)/(0.082061$423)
n = 0.508 moles H2
Step 4: Calculate moles NO
p*V = n*R*T
⇒ with p = the pressure of NO = 602 torr = 602/760 = 0.792165 atm
⇒ with V = the volume = 25.0 L
⇒ with n = the number of moles of NO = TO BE DETERMINED
⇒ with R= the gas constant = 0.08206 L*atm/mol*K
⇒ with T =the temperature = 150.0 °C = 423 K
n= (0.792165*25.0)/(0.08206*423)
n = 0.571 moles NO
Step 5: Calculate the limiting reactant
For 3 moles of H2 we need 2 moles of NO to produce 2 moles of NH3 and 1 mole of O2
H2 is the limiting reactant. It will be completely consumed (0.508 moles).
NO is in excess.
Step 6: Calculate moles NH3
For 3 moles of H2 we need 2 moles of NO to produce 2 moles of NH3
For 0.508 moles H2 we'll have 0.339 moles NH3 (This is when the reaction yield is 100%)
For reaction yield of 93.5 % the number of moles = 0.935 * 0.339 = 0.316965 moles NH3
Step 7: Calculate mass of NH3
Mass NH3 = moles NH3 * Molar mass NH3
Mass NH3 = 0.316965 moles NH3 * 17.03 5.398 grams
There will be produced 5.398 grams of NH3
