Question

For the following reaction, 0.308 grams of hydrogen gas are allowed to react with 43.2 grams of iodine. hydrogen(g)+ iodine(s) yields hydrogen iodide(g) What is the maximum mass of hydrogen iodide that can be formed? ______? What is the formula for limiti

Answer 1

I'm assuming you're referring to this reaction,

`H_2(g) + I_2(g) rightleftharpoons 2HI(g)`

Given,

`n_(I_2) = 43.2"g" * "mol"/(253.8"g") approx 0.170"mol"`

`n_(H_2) = 0.308"g" * "mol"/(2.02"g") approx 0.152"mol"`

We conclude hydrogen gas is the limiting reactant.

Hence,

`0.152"mol" * (2HI)/(H_2) * (127.9"g")/(HI) approx 3.90"g"`

is the maximum mass of hydrogen iodide gas able to be produced.