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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

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kenmyna
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: The correct answer is 1.32 mol [tex]H_2[/tex]

Explanation:

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

[tex]2Li+2HF\rightarrow 2LiF+H_2[/tex]

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 = [tex]\frac{1}{2}\times 3.5=1.75mol[/tex] of hydrogen gas.

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]   ...(1)

Moles of hydrogen gas = 1.75 mol

Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

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

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

[tex]\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100[/tex]

Percentage yield = 75.5 %

Theoretical yield = 3.5 g

Putting values in above equation, we get:

[tex]75.5=\frac{\text{actual yield}}{3.5g}\times 100\\\\\text{Actual yield}=2.64g[/tex]

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

[tex]\text{Moles of hydrogen gas produced}=\frac{2.64g}{2g/mol}=1.32mol[/tex]

Hence, the correct answer is 1.32 mol [tex]H_2[/tex]