Question

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.391 g of carbon dioxide is produced from the reaction of 0.16 g of methane and 0.84 g of oxygen gas, calculate the percent yield of carbon dioxide.

Answer 1

Answer : The percent yield of
`CO_2` is, 68.4 %

Solution : Given,

Mass of
`CH_4` = 0.16 g

Mass of
`O_2` = 0.84 g

Molar mass of
`CH_4` = 16 g/mole

Molar mass of
`O_2` = 32 g/mole

Molar mass of
`CO_2` = 44 g/mole

First we have to calculate the moles of
`CH_4` and
`O_2`.

`\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=(0.16g)/(16g/mole)=0.01moles`

`\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(0.84g)/(32g/mole)=0.026moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`CH_4+2O_2\rightarrow CO_2+2H_2O`

From the balanced reaction we conclude that

As, 2 mole of
`O_2` react with 1 mole of
`CH_4`

So, 0.026 moles of
`O_2` react with
`(0.026)/(2)=0.013` moles of
`CH_4`

From this we conclude that,
`CH_4` is an excess reagent because the given moles are greater than the required moles and
`O_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`CO_2`

From the reaction, we conclude that

As, 2 mole of
`O_2` react to give 1 mole of
`CO_2`

So, 0.026 moles of
`O_2` react to give
`(0.026)/(2)=0.013` moles of
`CO_2`

Now we have to calculate the mass of
`CO_2`

`\text{ Mass of }CO_2=\text{ Moles of }CO_2* \text{ Molar mass of }CO_2`

`\text{ Mass of }CO_2=(0.013moles)* (44g/mole)=0.572g`

Theoretical yield of
`CO_2` = 0.572 g

Experimental yield of
`CO_2` = 0.391 g

Now we have to calculate the percent yield of
`CO_2`

`\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}* 100`

`\% \text{ yield of }CO_2=(0.391g)/(0.572g)* 100=68.4\%`

Therefore, the percent yield of
`CO_2` is, 68.4 %