Question

3000 grams of heptane is combusted with 25000 grams of oxygen. C7H16 + O2 --> CO2 + H2O a) What is the limiting reactant? b) How many grams of carbon dioxide is produced? c) How many grams of excess reactant are left?

Answer 1

Heptane is limiting while oxygen is in excess. The mass of carbon dioxide produced would be 9,223.62 grams, while excess oxygen would measure 14,461.12 grams

Stoichiometric reaction

From the equation of the reaction:

C7H16 + 11O2 --> 7CO2 + 8H2O

Mole ratio of C7H16 to O2 = 1:11

Mole of 3000 g heptane = 3000/100.21

                                           = 29.94 moles

Mole of 25000 g O2 = 25000/32

                                    = 781.25 moles

Thus, O2 is in excess and heptane is limiting.

Mole ratio of heptane and CO2 = 1:7

equivalent mole of CO2 = 29.94 x 7 = 209.58 moles

Mass of 209.58 moles CO2 = 209.58 x 44.01

                                               = 9223.62 grams

Excess mole of O2 = 781.25 - (29.94x11)

                                = 781.25 - 329.34 = 451.91 moles

Mass of 451.91 moles of O2 = 451.91 x 32

                                              = 14,461.12 grams excess oxygen.

More on stoichiometric calculations can be found here: https://brainly.com/question/8062886