Consider the reaction at 500 ° c 500°c . n 2 ( g ) + 3 h 2 ( g ) − ⇀ ↽ − 2 nh 3 ( g ) k c = 0.061 n2(g)+3h2(g)↽−−⇀2nh3(g)kc=0.061 if analysis shows that the composition of the reaction mixture at 500 ° c 500°c is 1.14 mol ⋅ l − 1 n 2 1.14 mol⋅ l−1n2 , 5.52 mol ⋅ l − 1 h 2 5.52 mol⋅ l−1h2 , and 3.42 mol ⋅ l − 1 nh 3 3.42 mol⋅ l−1nh3 , what is the value of the reaction quotient q q ?

Q = 0.061 = Kc

Explanation:

Step 1: Data given

Temperature = 500 °C

Kc=0.061

1.14 mol/L  N2

5.52 mol/L H2

3.42 mol/L NH3

Step 2: Calculate Q

Q=[products]/[reactants]=[NH3]²/ [N2][H2]³

If Qc=Kc then the reaction is at equilibrium.  

If Qc<Kc then the reaction will shift right to reach equilibrium.

If Qc>Kc then the reaction will shift left to reach equilibrium.  

Q = (3.42)² / (1.14 * 5.52³)

Q = 11.6964/191.744

Q = 0.061

Q = Kc the reaction is at equilibrium.  

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g