Question

Propane (C3H8) burns in a combustion reaction. How many grams of C3H8 are needed to produce 80.3 mols CO2 ?

Answer 1

Answer:

1177.88g of C3H8

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Next we shall determine the number of mole of C3H8 required to produce 80.3 moles of CO2. This is illustrated below:

From the balanced equation above,

1 mole of C3H8 reacted to produce 3 moles of CO2.

Therefore, Xmol of C3H8 will react to produce 80.3 moles of CO2 i.e

Xmol of C3H8 = 80.3/3

Xmol of C3H8 = 26.77 moles

Finally, we shall convert 26.77 moles of C3H8 to grams.

Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

Mole of C3H8 = 26.77 moles

Mass of C3H8 =..?

Mass = mole x molar mass

Mass of C3H8 = 26.77 x 44

Mass of C3H8 = 1177.88g

Therefore, 1177.88g of C3H8 are needed for the reaction