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?
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
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