Question

If the reaction of 3.50 moles of lithium with excess hydrofluoric acid produced a 75.5% yield of hydrogen gas, what was the actual yield of hydrogen gas? unbalanced equation: li + hf lif + h2 1.32 mol h2 1.75 mol h2 2.32 mol h2 2.65 mol h2

Answer 1

balanced chemical equation is as follows
2Li +2HF --->2 Li + H2
from reaction above 2 moles of Li react with 2 moles of HF to form 2 moles of Li and 1 mole of H2
we know the mole of Li is 3.50 moles
Since ratio of Li to H2 is 2:1 then the mole of H2 is 1/2 x3.50=1.75moles

Answer 2

Answer: The correct answer is 1.32 mol
`H_2`

Step-by-step explanation:

For the reaction of lithium and hydrofluoric acid, the equation follows:

`2Li+2HF\rightarrow 2LiF+H_2`

By Stoichiometry of the reaction:

if 2 moles of lithium is producing 1 mole of hydrogen gas,

Then, 3.50 moles of lithium will produce =
`(1)/(2)* 3.5=1.75mol` of hydrogen gas.

• Now, to know the theoretical yield of hydrogen gas, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ...(1)

Moles of hydrogen gas = 1.75 mol

Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

`1.75mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Given mass of hydrogen gas}=3.5gl`

• To calculate the percentage yield, we use the equation:

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

Percentage yield = 75.5 %

Theoretical yield = 3.5 g

Putting values in above equation, we get:

`75.5=\frac{\text{actual yield}}{3.5g}* 100\\\\\text{Actual yield}=2.64g`

• Now, calculating the moles of hydrogen gas, we put the value in equation 1, we get:

`\text{Moles of hydrogen gas produced}=(2.64g)/(2g/mol)=1.32mol`

Hence, the correct answer is 1.32 mol
`H_2`