Question

For the following reaction, 10.1 grams of propane (C3H8) are allowed to react with 20.6 grams of oxygen gas . propane (C3H8)(g) + oxygen(g) carbon dioxide(g) + water(g) What is the maximum amount of carbon dioxide that can be formed? _ grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? _ grams

Answer 1

The balanced chemical equation would be :

C3H8 + 5 O2 =====> 3CO2 + 4H2O

mol of C3H8 in the reaction = mass / (molar mass x stoichiometric coefficient)

= 10.1 g / (44.1 gmol-1 x 1)

= 0.23 mol

mol of O2 = 20.6 g / ( 32 gmol x 5)

= 0.128 mol

Clearly, mol of O2 are the least, so it would be the limiting reagent.

So, formula of limiting reagent :O2

a) theoratical yield of CO2 = (mass of limiting reagent / molar mass of limiting reagent) x (stoichiometric coefficient of CO2 / stoichiometric coefficient of Limiting reagent) x molar mass of CO2

= (20.6 g / 32 gmol-1) x (3/5) x 44.01gmol-1

= 16.9 g

Now, since

O2 = limiting reagent

C3H8 =excess reagent

So,

Mass of excess reagent consumed = (mass of limiting reagent / molar mass of limiting reagent) x (stoichiometric coefficient of excess reagent / stoichiometric coefficient of Limiting reagent) x molar mass of excess reagent

= (20.6 / 32gmol-1) x (1 / 5) x 44.1gmol-1

= 5.67 g of C3H8 consumed

So, excess reactant left over = total mass - mass consumed

= (10.1 - 5.67) g

= 4.42 g