Question

For the chemical reaction below, determine the amount of hi produced when 3.35 g of hydrogen is reacted with 50.75 g of iodine to produce hydrogen iodide

Answer 1

Molar mass of H₂ = 1.008 × 2 g/mol = 2.016 g/mol 
Molar mass of I₂ = 126.9 × 2 g/mol = 253.8 g/mol 
Molar mass of HI = (1.008 + 126.9) g/mol = 127.9 g/mol 

H₂(g) + I₂(g) → 2HI 
Mole ratio H₂ : I₂ : HI = 1 : 1 : 2 

Then the initial number of moles of H₂ = (3.35 g) / (2.016 g/mol) = 1.662 mol 
Initial number of moles of I₂ = (50.75 g) / (253.8 g/mol) = 0.2000 mol < 1.662 mol 
Hence, I₂ is the limiting reactant (limiting reagent). 

Number of moles of I₂ reacted = 0.2000 mol 
Number of moles of HI reacted = (0.2000 mol) × 2 = 0.4000 mol 
Mass of HI reacted = (127.9 g/mol) × (0.4000 mol) = 51.16 g

Answer 2

The amount of HI produced when 3.35 g of hydrogen is reacted with 50.75 g of iodine will be 51.16 g.

What is a chemical reaction?

One or more compounds, known as reactants, are transformed into one or more distinct substances, known as products. Such a process is known as a chemical reaction.

The standard balanced chemical equation for the given condition;

H₂(g) + I₂(g) → 2HI

Mole ratio H₂ : I₂: HI is  1 : 1: 2.

The molar mass of H₂ is;

$1080 \times 2 =2.016 \ g/mol$

The Molar mass of I₂ is;

$1269 \times 2 =253.8 \ g/mol$

The molar mass of HI is;

$(1.008+126.9) =127.9\ g/mol$

The no of mole for the hydrogen before the reaction;

$n_H =\frac{335}{2.0016} =1.662 \ mol$

The no of mole for the I₂  before the reaction;

$\rm n_I= \frac{50.75}{253.8 } \\\\ \rm n_I= 0.2000$

The no of a mole of the I₂ is less. So that it will be a limiting reagent.

The no of a mole of HI reacted;

$N= \frac{127.9}{0.4000} \\\\ N=51.16 g$

Hence the amount of HI produced when 3.35 g of hydrogen is reacted with 50.75 g of iodine will be 51.16 g.

To learn more about the chemical reaction refer to the link;

https://brainly.com/question/22817140