Question

If the reaction of 3.75 moles of lithium with excess hydrofluoric acid produced a 96.5% yield of hydrogen gas, what was the actual yield of hydrogen gas? unbalanced equation: li + hfyields lif + h2

Answer 1

1.81 mol H2. Hope this answer helped!

Answer 2

Answer : The actual yield of hydrogen gas is, 3.618 grams

Explanation : Given,

Given moles of
`Li` = 3.75 moles

Molar mass of
`H_2` = 2 g/mole

Now we have to calculate the moles of
`H_2`.

The balanced chemical reaction will be,

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

From the given balanced reaction, we conclude that

As, 2 moles of
`Li` react to give 1 mole of
`H_2`

So, 3.75 moles of
`Li` react to give
`(3.75)/(2)=1.875` moles of
`H_2`

Now we have to calculate the mass of
`H_2`.

`\text{Mass of }H_2=\text{Moles of }H_2* \text{Molar mass of }H_2`

`\text{Mass of }H_2=(1.875mole)* (2g/mole)=3.75g`

The theoretical yield of
`H_2` is, 3.75 grams

Now we have to calculate the actual yield of
`H_2`.

`\%\text{ yield of }H_2=\frac{\text{Actual yield of }H_2}{\text{Theoretical yield of }H_2}* 100`

`96.5=\frac{\text{Actual yield of }H_2}{3.75g}* 100`

`\text{Actual yield of }H_2=3.618g`

Therefore, the actual yield of hydrogen gas is, 3.618 grams