Question

Consider the following reaction. C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) If a container were to have 28 molecules of C5H12 and 28 molecules of O2 initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion? (Use only whole numbers to solve.)

Answer 1

according to balanced reaction

8 moles O2 reacts with 1 mole C5H12

28 moles O2 reacts with 28 x 1 / 8 = 3.5 moles C5H12

moles of C5H12 left after reaction = 28 - 3.5 = 24.5

8 moles O2 formed 5 moles of CO2

28 moles O2 forms 28 x 5 / 8 = 17.5 moles CO2

8 moles O2 formed 6 moles H2O

28 moles O2 forms 28 x 6 / 8 = 21 moles H2O

total moles after reaction complete = 21 moles H2O + 17.5 moles CO2 + 24.5 moles C5H12 = 63

1 mole = 6.023 x 10²³ molecules

63 moles = 63 x 6.023 x 10²³/1 = 3.80 x 10²⁵ molecules

answer = 3.80 x 10²⁵ molecules