Question

11) 2NH3(g)+3O2(g)+2CH4(g) --->2HCN(g) +6H2O(g)

^=-45.9 kj ^=-74.87kj ^= 135.1kj ^= -241.8kj

a) Calculate delta H rxn

b) Calc q when 25.0 g CH4 reacts

c) If 10.0L of each reactant was used at STP what V of HCN would be formed?

d) If 6.0L of HCN was formed? calc % yield

Answer 1

2NH3(g)+3O2(g)+2CH4(g) --->2HCN(g) +6H2O(g) : H = ?

(a) H = H^o_f products - H^o_f reactants

           = [(2x H^o_f HCN(g)) + (6x H^o_f H2O(g))] - [(2x H^o_{f NH3(g)) + (3x}H^o_f O2(g)) +(2x H^o_f CH4(g))]

           = [(2x135.1)+(6x(-241.8)]-[(2x(-45.9))+(3x0)+(2x(-74.87))]

          = -939.1 kJ

(b) From the balanced reaction,

2 mol = 2x16 g of CH4 produces 939.1 kJ of heat

25.0 g of CH4 produces M kJ of heat

M = ( 25.0 x 939.1)/(2x16)

   = 733.6 kJ

Therefore the heat released or amount of heat , q = -733.6 kJ

(c) We know that as Temperature & Pressure kept constant , Volume is proportional to number of moles

From the balanced reaction ,

3 moles of O2 produces 2 moles of HCN

OR 3L of O2 produces 2 L of HCN

     10 L of O2 produces X L of HCN

     X = ( 10x2) / 3 = 6.67 L of HCN formed

(d) Percent yield = [6.0)/6.67]x100

                        = 89.9 %