Question

If a reaction of 5.0 g of hydrogen with 5.0 g of carbon monoxide produced 4.5 g of methanol, what was the percent yield?

Answer 1

Answer:

The percentage yield of methanol is 78.74%.

Explanation:

$2H_2+CO\rightarrow CH_3OH$

Theoretical yield of methanol ;

Moles of hydrogen gas = $\frac{5.0 g}{2 g/mol}=2.5 mol$

Moles of carbon monoxide = $\frac{5.0 g}{28 g/mol}=0.1786 mol$

According to reaction ,1 mole of CO reacts with 2 moles of hydrogen gas. Then 0.1786 moles of CO will :

$\frac{2}{1}\times 0.1786 mol=0.0893 mol$

As we can see, that moles of hydrogen gas are in excess and CO are in limiting amount, so amount of methanol will depend upon moles of CO.

According to recation 1 mole of CO gives 1 mole of methanol, then 0.1786 moles of CO will give:

$\frac{1}{1}\times 0.1786 mol=0.1786 mol$

Mass of 0.1786 moles of methanol =

= 0.1786 mol × 32 g/mol = 5.7152 g

Theoretical yield of methanol  = 5.7152 g

Experimental yield of methanol = 4.5 g

Percentage yield of methanol:

To calculate the percentage yield , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{4.5 g}{5.7152 g}\times 100=78.74\%$

The percentage yield of methanol is 78.74%.