Question

For the chemical reaction below, determine the amount of HI produced when 3.32E+0 g of hydrogen is reacted with 5.064E+1 g of iodine to produce hydrogen iodide (HI). H(g) + I(g) → 2HI(g)

Answer 1

Answer:  The amount of HI produced is 102 g

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

For hydrogen:

Given mass of hydrogen = $3.32\times 10^0g=3.32g$

Molar mass of hydrogen= 2 g/mol

$\text{Moles of hydrogen}=\frac{3.32g}{2g/mol}=1.66mol$

For iodine:

Given mass of iodine =  $5.064\times 10^1g=50.64g$

Molar mass of iodine= 127 g/mol

$\text{Moles of iodine}=\frac{50.64g}{127g/mol}=0.399mol$

The chemical equation for the reaction is:

$H_2(g)+I_2\rightarrow 2HI(g)$

By Stoichiometry of the reaction:

1 mole of iodine reacts with 1 mole of hydrogen

So, 0.399 moles of  iodine will react with = $\frac{1}{1}\times 0.399=0.399mol$ of hydrogen

As, given amount of hydrogen is more than the required amount. So, it is considered as an excess reagent.

Thus, iodine is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 moles of iodine produces 2 mole of hydrogen iodide

So, 0.399 moles of iodine will produce = $\frac{2}{1}\times 0.399=0.798moles$ of hydrogen iodide

$0.798mol=\frac{\text{Mass of hydrogen iodide }}{128g/mol}\\\\\text{Mass of hydrogen iodide}=102g$

The amount of HI produced is 102 g