Question

22.4 g of pentane (C5H12) and 17.0 g of O2 are available for a combustion reaction. If you actually produce 1.0 g of water, what is the percent yield?

Answer 1

`2.98\text{ \%}`

Step-by-step explanation:

Here, we want to get the percentage yield

Mathematically, we have that as:

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

We have the actual yield, now let us calculate the theoretical yield

We start by writing the complete equation of the reaction

We have that as:

`\text{ C}_5H_(12)\text{ + 8O}_2\rightarrow\text{ 5CO}_2+6H_2O`

In a combustion reaction, the hydrocarbon which in this case methane is usually the limiting reactant

Thus, we can calculate the mass of water produced by using the mass of Pentane given

Firstly, we need to get the number of moles of pentane that reacted

To get that, we have to divide the mass of pentane given by the molar mass of pentane

The molar mass of pentane is 72 g/mol

Thus, we have the number of moles as:

`(22.4)/(72)\text{ = 0.31 mole}`

From the equation of reaction:

1 mole of pentane produced 6 moles of water

0.311 mole of pentane will produce:

`6\text{ }*\text{ 0.311 = 1.87 moles water}`

Now, to get the mass of water produced

We multiply this number of moles by the molar mass of water

The molar mass of water is 18 g/mol

That means the mass theoretically produced will be:

`1.87\text{ }*\text{ 18 = 33.60 g}`

From here, we have the percentage yield as:

`\text{ \% yield = }(1)/(33.6)*\text{ 100 \% = 2.98\%}`