Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 8.0 g of octane is mixed with 44.0 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Question
Asked Dec 3, 2019
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Liquid octane 

CH3CH26CH3

 will react with gaseous oxygen 

O2

 to produce gaseous carbon dioxide 

CO2

 and gaseous water 

H2O

. Suppose 8.0 g of octane is mixed with 44.0 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

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Expert Answer

Step 1

Given that 8g of octane and 44 gO2 are mixed in a reaction. Number of moles of C8H18 and O2 can be determined:

2C,H1+250216CO2+18H,0
mass
No.of moles of C,H1
molar mass
8g
114g/mol
0.07mol
mass
No.of moles of O,
molar mass
44g
32g/mol
-1.375mol
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2C,H1+250216CO2+18H,0 mass No.of moles of C,H1 molar mass 8g 114g/mol 0.07mol mass No.of moles of O, molar mass 44g 32g/mol -1.375mol

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Step 2

Limiting reagent can ...

2C,H+250216CO2+18H,0
2molC,H reacts ->25mol02
18
0.07molC,H reacts»x molO2
25molO2 x0.07 mol C,H
2mol C,H18
No.of moles of O2
0.875mol
No.of moles of O,present =1.375mol
Therefore, C3H18 will be the limiting reagent
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2C,H+250216CO2+18H,0 2molC,H reacts ->25mol02 18 0.07molC,H reacts»x molO2 25molO2 x0.07 mol C,H 2mol C,H18 No.of moles of O2 0.875mol No.of moles of O,present =1.375mol Therefore, C3H18 will be the limiting reagent

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