Question

A sample of propane (20.4 g) is burned in 82.3 g of oxygen in a chemical reaction to produce water and carbon dioxide (CO₂). If 75.5 g of CO₂ is formed, how much water, in grams, was also produced?.

How I can get the answer 27.2.

Answer 1

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

1 mol 5 mol

44 g 5 x32 = 160 g

20.4 g 82.3 g

First we have to identify the limiting reagent.

  C3H8 :

Given mass of C3H8 = 20.4 g

         molar mass of C3H8 from balanced equation = 44 gmol-1

         moles of C3H8 = mass/molar mass = 20.4 g/ 44 g mol-1

                                                            = 0.464 mol

         moles of C3H8 = 0.464 mol

O2 :

          given mass of O2 = 82.3 g

        molar mass of O2 from balanced equation = 160 gmol-1

         moles of O2 = mass/molar mass = 82.3 g/ 160 g mol-1

                                                            = 0.51 mol

     Hence, moles of O2 = 0.51 mol

                moles of C3H8 = 0.464 mol

Here C3H8 present in lesser no of moles. Therefore, C3H8 is the limiting reagent and yield is calculated based on C3H8.

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

1 mol 4 mol

44 g 4 x18 = 72 g   

20.4 g ?

? = (20.4 g/ 44 g) x 72 g water

= 33.3 g water