Question

53, Assuming the reaction below goes to comp letfon, H02 4NO2 6H20 hON many muleau Zes of water are produced when 0.0170 gram of ammonia reacts with -3 1.57 x 10 liter of oxygen gas at STP? (1) 2.41 x 109 (2) 3,62 x 1019 (3) 4.21 x 109 (4) 6.02 x 10 19 54, Analysis shows a compound to contain 18.81% carbo

Answer 1

Ans. # Step 1: Determine the limiting reactant:

In the balanced reaction-

            Theoretical molar ratio of reactant:           NH₃ : O₂ = 4 : 7

# Moles of NH₃ = Mass / MW = 0.0170 g / (17.03 g/ mol) = 9.982 x 10^-4 mol

# Calculate moles of O₂ using ideal gas equation.

Equation 1 Pressure in atm Ideal Gas equation : PV = nRT where, P = V = Volume in liters nNumber of moles R = 0.0821 atm L mol-1 K-1 T = Temperature in kelvin GiVen, V1.57E-03 L 0.000C= 273.15 K 1.000 atnm Putting the values in equation 1 1 atm X 1.57E-03 L 0.0821 atm L mol1 K1x 273.15 K Hence, n 7.001E-05 mol

# Experimental molar ratio of reactant:     NH₃ : O₂ = 9.982 x 10^-4 mol : 7.001 x 10^-5 mol

                                                            = 4 : 0.28

# Comparing the theoretical and experimental molar ratios of reactants, the experimental moles of O₂ is less than its theoretical value 7 while keeping that of NH₃ constant at 4 mol. Therefore, O₂ is the limiting reactant.

# Step 2: Calculate the yield:

The formation of product follows the stoichiometry of limiting reactant.

Following stoichiometry of balanced reaction, 7 mol O₂ produces 6 mol H₂O.

So,

            Moles of H₂O formed = (6 / 7) x moles of O₂ consumed

                                                = (6 / 7) x 7.001 x 10^-5 mol

                                                = 6.000 x 10^-5 mol

Now,

            Molecules of H₂O formed = Moles x N_A

                                                = 6.00 x 10^-5 mol x (6.022 x 10²³ molecules/ mol)

                                                = 3.6137 x 10¹⁹ molecules.

So, correct option is- 2. 6.32 x 10¹⁹ (closest value)