Hydrogen iodide can decompose into hydrogen and iodine gases. 2HI(g) H2(g) I2(g) K for the reaction is 0.016. Of 0.148 atm of HI(g) is sealed in a flask, what is the pressure of each gas when equilibrium is established

Solution :

Given :

Hydrogen iodide decomposes to hydrogen and iodine gas

I    0.148            0          0

C     -2a           +a          +a

E  0.148-2a      a            a

We know

0.0187 = 1.2529 a

a = 0.0149

Therefore

P(HI) = 0.148 - 2a

        = 0.148 - 2(0.0149)

        = 0.1182 atm

P() =  a

        = 0.0149 atm

P() =  a

        = 0.0149 atm

law of conservation of matter

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