1.

How many grams of oxygen are required to burn completely 570g of octane?

Answer» `underset(2xx114)underset(2"mol")(2C_(8)H_(18))+underset(25xx32)underset("25 mole")(25O_(2))to16CO_(2)+18H_(2)O`
First method: For burning `2xx114g` of octane, oxygen required `=25xx32g`
Thus, for burning570 g of octane, oxygen required `(25xx32)/(2xx114)xx570g=2000g`
Molar method: Number of moles of octane in 570grams `=(570)/(114)=5.0`
For burning 2.0 moles of octane, oxygen required `=25"mol"=25xx32g`
For buring 5 moles of octane, oxygen required `=(25xx32)/(2.0)xx5.0g=2000g`
Proportion method: Let x g of oxygen be required for burning 570.0g of octane. It is known that `2xx114g` of teh octane require `25xx32g` of oxygen, then, the proportion,
`(25xx32"g oxygen")/(2xx114"g octane")=(x)/("570.0g octane")`
`therefore x=(25xx32xx570)/(2xx114)=2000g`


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