Question

In a laboratory experiment the reaction of 3.0 mol of H2 with 2.0 mol of I2 produced 1.0 mole of HI. Determine theoratical yield in grams and the percent yield for this reaction.

Answer 1

Theoretical yield of HI is 512 g.

The percent yield for this reaction is 25%.

Step-by-step explanation:

`H_2+I_2\rightarrow 2HI`

Moles of hydrogen gas = 3.0 moles

Moles of iodine gas = 2.0 moles

According to reaction 1 mol of hydrogen gas reacts with 1 mol of iodine gas.

Then 3.0 moles of hydrogen gas reacts with 3.0 mol of iodine gas. But there are 2.0 moles of iodine gas. Hence,Iodine is a limiting reagent. The production of HI will depend upon iodine gas moles.

According to reaction , 1 mol of iodine gas gives 2 moles of HI.

Then 2 moles of iodine gas will give:

`(2)/(1)* 2 mol=2 mol` of HI

Theoretically we will get 4 moles of HI.

Theoretical yield of HI = 4 mol × 128 g/mol= 512 g

Experimental yield of HI = 1.0 mol

= 1 mol × 128 g/mol= 128 g

`\%yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}} * 100`

`\%yield=(128 g)/(512 g)* 100=25\%`

The percent yield for this reaction is 25%.