Question

Volume of CO2 gas obtained at STP when 9 g oxalic acid is
treated with conc. H2SO4 as per the following reaction is
(COOH)2
conc. Hą504_CO2 + CO + H20
4.48 L
1.12 L
2.24 L​

Answer 1

Answer:

1.06 g of sodium carbonate correspond to

106

1.06

=0.01 moles.

Na

2

CO

3

+2HCl→2NaCl+H

2

O+CO

2

1 mole of sodium carbonate liberates 1 mole of CO

2

.

0.01 moles of sodium carbonate will liberate 0.01 moles of CO

2

.

1 mole of CO

2

at STP occupies a volume of 22400 cm

3

.

0.01 mole of CO

2

at STP will occupy a volume of 224 cm

3

.

Explanation:

224cm ^3

Answer 2

Given - Mass of oxalic acid : 9 gram

Find - Volume of CO2 gas obtained

Solution - The mentioned reaction of oxalic acid is as follows -

(COOH)2 + conc. H2S04 --> CO2 + CO + H20

As per the reaction, 1 mole of oxalic acid releases 22.4 litre of carbon dioxide gas at STP.

The mass of 1 mole of oxalic acid is 90 gram.

Hence, 90 gram oxalic acid releases 22.4 litre of carbon dioxide.

So, 9 gram of oxalic acid will release = 22.4/90*9

9 gram of oxalic acid will release = 2.24 litre.

Thus, correct option is 2.24 L.