Question

Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.

percent yield = 0.8347g / 0.9525 g × 100% = 87.6%

Answer 1

Answer : The limiting reactant is
`O_2`

Explanation : Given,

Mass of
`C_3H_8` = 30.0 g

Mass of
`O_2` = 75.0 g

Molar mass of
`C_3H_8` = 44 g/mole

Molar mass of
`O_2` = 32 g/mole

First we have to calculate the moles of
`C_3H_8` and
`O_2`.

`\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=(30.0g)/(44g/mole)=0.682moles`

`\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(75.0g)/(32g/mole)=2.34moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`C_3H_8+5O_2\rightarrow 3CO_2+4H_2O`

From the balanced reaction we conclude that

As, 5 mole of
`O_2` react with 1 mole of
`C_3H_8`

So, 2.34 moles of
`O_2` react with
`(2.34)/(5)* 1=0.468` moles of
`C_3H_8`

From this we conclude that,
`C_3H_8` 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.

Therefore, the limiting reactant is
`O_2`