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)

 The amount of HI produced is 102 g

Explanation:

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

     .....(1)

For hydrogen:

Given mass of hydrogen =

Molar mass of hydrogen= 2 g/mol

For iodine:

Given mass of iodine =  

Molar mass of iodine= 127 g/mol

The chemical equation for the reaction is:

By Stoichiometry of the reaction:

1 mole of iodine reacts with 1 mole of hydrogen

So, 0.399 moles of  iodine will react with = 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 = of hydrogen iodide

The amount of HI produced is 102 g

i wish i could but this one is weird give me abc and i can solve it.