Cal the vol of carbon(ii)oxide produced at stp by. d dehydration of 6.00g of methanoic acid with concentrated H2SO4 carbon=2 oxygen=16 hydrogen=1 molar volume = 22.4dmcube

The volume of CO= 2.912 L

Given

Dehydration of 6 g of methanoic acid with H2SO4

Required

The volume of CO

Solution

Reaction

HCOOH  ⇒  CO+H₂O

MW = 1.Ar C+2.Ar H+2.Ar O

MW = 1.12+2.1+2.16

MW= 46 g/mol

mol = mass : MW

mol = 6 g : 46 g/mol

mol = 0.13

From equation, mol ratio HCOOH  : CO, so mol CO = 0.13

Vm = 22.4 L

so for 0.13 mol :

= 0.13 x 22.4 L

= 2.912 L

false

because it does not have the potential to be entirely used up

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