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

Step-by-step 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* 10^0g=3.32g`

Molar mass of hydrogen= 2 g/mol

`\text{Moles of hydrogen}=(3.32g)/(2g/mol)=1.66mol`

For iodine:

Given mass of iodine =
`5.064* 10^1g=50.64g`

Molar mass of iodine= 127 g/mol

`\text{Moles of iodine}=(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 =
`(1)/(1)* 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 =
`(2)/(1)* 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