Question

Liquid hexane(CH₃ (CH₂)₄CH₃)will react with gaseous oxygen(O₂ )to produce gaseous carbon dioxide (CO₂)and gaseous water(H₂O). Suppose 29. g of hexane is mixed with 158. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

first write the balanced equation and i waas hoping that you know how to balance the reaction

2CH3 (CH2)4CH3 + 19O2 -----> 12CO2 + 14H2O

you have started with 29.0 grams of hexane calculate the moles from it

no of moles of hexane = weight of hexane / molar mass f hexane

= 29.0 / 86.11 (molar mass of hexane taken from google)

= 0.3367 moles of hexane

no of moles of O2

weight of O2 / molar mass of O2

= 158 / 32

= 4.94 moles

so here limiting agent is hexane

from the balanced equation it is clear that

14 moles of H2O will produce from the the 2 moles of hexane

how many moles of H2O will produce from the 0.3367 moles of hexane

use cross multiplication

= 14 x 0.3367 / 2

= 0.6734 moles of H2O will produce

now we know the moles from this we can calculate the weight of H2O

weight of H2O = moles of H2O x molar mass of H2O

= 0.6734 x 18.01

= 0.0374 grams

= 0.04 grams