One litre of oxygen at NTP weighs 1.46 g. How many litres of oxygen are needed for the combustion of `21.0` g of Mg whose equivalent weight is `1/2` mole?

Explanation

The volume occupied by 1 equivalent at NTP is equivalent volume.
`therefore` equivalent volume of oxygen `=` volume of 8.0 g of oxygen at NTP (eq. wt. of O = 8 )
`= ( 8)/( 1.46) = 5.48` litres. (1.46) g oxygen contains 1 L at NTP)
Given that the equivalent weight of `Mg = (1)/(2) xx 24 = 12 " "`(wt. of 1 mole of Mg = 24 g)
Now, in the combustion,
equivalent of Mg = equivalent of oxygen.
`therefore ("weight of Mg")/("eq. weight of Mg")=("volume of oxygen at NTP")/("eq. volume of oxygen")`
`(21)/(12) = ("volume of oxygen")/(5.48)`
`therefore` volume of oxygen `=` 9.59 litres.