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 : The limiting reactant is

Explanation : Given,

Mass of = 30.0 g

Mass of = 75.0 g

Molar mass of = 44 g/mole

Molar mass of = 32 g/mole

First we have to calculate the moles of and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 5 mole of react with 1 mole of

So, 2.34 moles of react with moles of

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

Therefore, the limiting reactant is

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