cal the vol of carbon(ii)oxide produced at stp by. d dehydration of 6.00g of methanoic acid with concentrated H2SO4 (carbon =2oxygen=16 hydrogen=1 molar volume = 22.4dmcube plse I really need it a lot

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

particle   p is an electron and particle q is a neutron

the answer is option a.