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 (2 points)

Answer 1

Answer:- the actual yield of hydrogen gas is 1.32 moles or 2.65 grams.

Solution:- The balanced equation is:

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

From this equation, there is 2:1 mol ratio between lithium and hydrogen gas. So, we could calculate the theoretical moles of the hydrogen gas formed.

$3.50molesLi(\frac{1molH_2}{2molLi})$

= 1.75 moles

The formula of percent yield is:

%yield = $(\frac{actual}{theoretical})100$

Using this formula the actual yield that is actual moles of hydrogen gas produced could be calculated.

$75.5=(\frac{actual}{1.75})100$

$actual=(\frac{75.5*1.75}{100})$

actual = 1.32 moles

So, the actual yield of hydrogen gas is 1.32 moles. If it wants answer in grams then multiply it by the molar mass(2.01 gram per mol) to get the answer in grams.

$1.32mol(\frac{2.01g}{1mol})$

= 2.65 g

Hence, the actual yield of hydrogen gas is 1.32 moles or 2.65 grams.