How many litres of `_2` at STP will be formed when 100 ml of `0.1` M,`H_2S)_4` reacts with excess of `Na_2CO_3` ?

Options

A. 22.4
B. 2.24
C. 0.224
D. 5.6

Answer

c

Explanation

Volume of `H_2SO_4=100ml`
Molarity=0.1 M
`therefore ``Molarity =("Numbers of moles of"H_(2)SO_(4))/("`therefore `olumes of "H_(2)SO_(4)(ml)x)xx1000`
`0.1=(" No of moles")/(100)xx1000``0.1=(" No of moles")/(100)xx1000`
` implies "No of moles"=(0.1xx100)/1000=0.01`
` thereforeunderset (("1 mole"))(H_2SO_4+Na_2CO_3)tounderset ((22.4L))(CO_2+Na_2SO_4+H_2O)`
`therefor` 1 mole of `H_2SO_4` reacts and gives 22.4 L of `CO_2`
`therefore` 0.01 mole of `H_2SO_4` reacts with`Na_2CO_3` to give
`=22.4xx0.01`
`=0.224` of `CO_2`