The combustion of hexane can be represented by the following equation: __C 6 H 1

Question

The combustion of hexane can be represented by the following equation:

__C6H14 (g) + __O2(g) ? __CO2(g) + __H2O(g)

Balance the equation and answer the next three questions.

If 256. g of C6H14 and 894. g O2 react, which reactant is used up first? (enter hexane or oxygen) oxygen

What is the mass of carbon dioxide produced? (Give your answer to 3 sig figs.) grams

What is the mass of water produced?(Give your answer to 3 sig figs.) grams

You are correct.
Your receipt no. is 155-27 Previous Tries The combustion of hexane can be represented by the following equation: Balance the equation and answer the next three questions. If 256. g of C6H14 and 894. g O2 react, which reactant is used up first? (enter hexane or oxygen) oxygen What is the mass of carbon dioxide produced? (Give your answer to 3 sig figs.) grams Calculate the number of moles of the limiting reagent. Use the stoichiometry of the balanced equation to determine how many moles of CO2 will be produced. Then convert this to grams of CO2 by using its molar mass. What is the mass of water produced?(Give your answer to 3 sig figs.) grams

Explanation / Answer

2C6H14(g) + 19O2(g) ? 12CO2(g) + 14H2O(g)

Molar mass of C6H14(g) = 86 g/mole

THus, moles of C6H14(g) in 256 g of it = 256/86 = 2.98

molar mass of O2(g) = 32 g/mole

Thus, moles of O2(g) in 894 g of it = 894/32 = 27.9375

Now, C6H14(g) & O2(g) reacts in the ratio of 2:19

Thus, for 2.98 moles of C6H14(g) moles of O2(g) required = 19*2.98/2 = 28.28

Clearly O2 is limiting reagent &  C6H14(g) is in excess of 0.04 moles

Hence excess amount of  C6H14(g) = 0.04*86 = 3.372 g

Moles of CO2 produced = 6*moles of  C6H14(g) reacting = 6*2.94 = 17.64

mass of CO2 produced = 17.64*44 = 776.16 g

Moles of H2O produced = 7*moles of  C6H14(g) reacting = 7*2.94 = 20.58

Mass of watre produced = 20.58*18 = 370.44 g

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