Respuesta :
Answer: The amount of HI produced is 102 g
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For hydrogen:
Given mass of hydrogen = [tex]3.32\times 10^0g=3.32g[/tex]
Molar mass of hydrogen= 2 g/mol
[tex]\text{Moles of hydrogen}=\frac{3.32g}{2g/mol}=1.66mol[/tex]
For iodine:
Given mass of iodine = [tex]5.064\times 10^1g=50.64g[/tex]
Molar mass of iodine= 127 g/mol
[tex]\text{Moles of iodine}=\frac{50.64g}{127g/mol}=0.399mol[/tex]
The chemical equation for the reaction is:
[tex]H_2(g)+I_2\rightarrow 2HI(g)[/tex]
By Stoichiometry of the reaction:
1 mole of iodine reacts with 1 mole of hydrogen
So, 0.399 moles of iodine will react with = [tex]\frac{1}{1}\times 0.399=0.399mol[/tex] 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 = [tex]\frac{2}{1}\times 0.399=0.798moles[/tex] of hydrogen iodide
[tex]0.798mol=\frac{\text{Mass of hydrogen iodide }}{128g/mol}\\\\\text{Mass of hydrogen iodide}=102g[/tex]
The amount of HI produced is 102 g
