Question

Consider the reaction:

H₂(g) + I₂(s)    =    2HI (g)

Given an initial mass of 19.02g H2 an excess of I₂ and assuming that all of the reactant is conserved to products and non is lost calculate the mass (g) of HI produced by the reaction?

Answer 1

Sol .

Reaction : H2(g) + I2(g) ---> 2HI(g)

As mass of H2 = 19.02 g

Molar mass of H2 = 2.02 g/mol

So , moles of H2 = mass of H2 / molar mass of H2 = 19.02 / 2.02 = 9.4158 mol

Now , from reaction , 1 mole of H2 produces 2 moles of HI .

So , 9.4158 moles of H2 produces = 9.4158*2 = 18.8316 moles of HI

As molar mass of HI = 127.91 g / mol

Therefore , mass of HI produced = Moles of HI produced * Molar mass of HI

= 18.8316*127.91

= 2408.75 g