Question

13.74 L of hydrogen gas at 30 degrees C and 801 torr and 6.55 L of oxygen gas at 25 degree C and 801 torr are drawn into a cylinder where the following reaction takes place.

2 H2 (g) + O2 (g) -> 2H2O (l)

what is the limiting reactant?
If 8.7 g of water are produced what is the percent yield?

Answer 1

We have an ideal gas equation PV = nRT => n = (PV)/RT 
R = Ideal Gas constant = 0.082
 T, temparature of hydroxen = 30 + 273.15 = 303.15K
 Temperature of oxygen = 25 + 273.15 = 298.15
 P, pressure is 801 torr for both oxygen and hydrogen, 801torr = 801mmHg 
 1 atm = 760mmHg => 801 torr = 1.054atm.
 1): For Hydrogen 13.74 L volume => 
 n = (1.054*13.74)/(0.082*303.15) = 0.582mol
 For Oxygen 6.55 L volume => 
 n = (1.054*6.55)/(0.082*298.15) = 0.282mol
 From equation there is 0.282mol of oxygen which is equal to the calculated
making it a limiting reactant whereas there is 0.291mol of hydrogen from
equation which is less than the calculated.
 Hence deducing that oxygen is the limiting reactant as it has very few moles.
 2): Calculating the yield of water with the help of stoichiometry principles and moles of oxygen, 
  = 0.282mol x (2molH2O/1molO2)*(18gH2O/1molH2O) = 36 x 0.282 g H2O
= 10.15gH2O
 Produced yield is 8.7 g of water. 
 Now calculating the percent yield = (actual x100)/ predicted = 8.7 x100/10.15
 The percent yield = 85.71%.