Question

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 94. g of octane is mixed with 161. g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

Answer:-

Given:-

given wt. of liquid octane (C₈H₁₈) = 94.0 g

wt. of oxygen (O₂) = 161.0 g

minimum mass of octane (C₈H₁₈) that could be left over by the chemical reaction (w) = ?

As we know that

molar mass of liquid octane (C₈H₁₈) = 8 molar mass of C + 18 molar mass of H

molar mass of liquid octane (C₈H₁₈) = 8 12 + 18 ​​​​​​​ 1

molar mass of liquid octane (C₈H₁₈) = 96 + 18

molar mass of liquid octane (C₈H₁₈) = 114 g / mol

similarly

molar mass of oxygen (O₂) = molar mass of O + molar mass of O

molar mass of oxygen (O₂) = 16 + 16

molar mass of oxygen (O₂) = 32 g / mol

similarly

molar mass of carbon dioxide (CO₂) = molar mass of C + 2 molar mass of O

molar mass of carbon dioxide (CO₂) = 12 + 2 ​​​​​​​ 16

molar mass of carbon dioxide (CO₂) = 12 + 32

molar mass of carbon dioxide (CO₂) = 44 g /mol

similarly

molar mass of water (H₂O) = 2 molar mass of H +molar mass of O

molar mass of water (H₂O) = 2 1 + 16

molar mass of water (H₂O) = 2 + 16

molar mass of water (H₂O) = 18 g /mol

Also we know that the balanced equation of Liquid octane (C₈H₁₈) with gaseous oxygen (O₂) is as follows:-

2C₈H_18(l) + 25O_2(g)    16CO_2(g) + 18H₂O_(g)

2114 g 2532 g 1644 g 1818 g

228 g 800 g 704 g 324 g

Since

800 g oxygen (O₂) reacts with = 228 g Liquid octane (C₈H₁₈)

1 g oxygen (O₂) reacts with = 228 / 800 g Liquid octane (C₈H₁₈)

then

161.0 g oxygen (O₂) reacts with = 228 161.0 / 800 g Liquid octane (C₈H₁₈)

161.0 g oxygen (O₂) reacts with = 36708‬.0 / 800 g Liquid octane (C₈H₁₈)

161.0 g oxygen (O₂) reacts with = 45.89 g Liquid octane (C₈H₁₈)

therefore 161.0 g oxygen (O₂) reacts with 45.89 g Liquid octane (C₈H₁₈) to produce gaseous carbon dioxide CO₂ and gaseous water H₂O.

mass of Liquid octane (C₈H₁₈) that reacts with 161.0 g oxygen (O₂) = 45.89 g

So

minimum mass of octane (C₈H₁₈) that could be left over by the chemical reaction (w) = given wt. of liquid octane (C₈H₁₈) -  mass of Liquid octane (C₈H₁₈) that reacts with 161.0 g oxygen (O₂)

minimum mass of octane (C₈H₁₈) that could be left over by the chemical reaction (w) = 94.0 g -  45.89 g

minimum mass of octane (C₈H₁₈) that could be left over by the chemical reaction (w) = 48.11 g (i.e the answer)