The volume of ${CO}_2$ formed when 1 liter of $O_2$ reacted with 2 lit of $CO$ under the same condition is

Given, amount of $O_2$ = 1 liter

Amount of CO = 2 liter

$CO+\frac{1}{2}{O}_2⟶{CO}_2$

1 mole of CO formed 1 mole of ${CO}_2$

2 liters of CO formed 2 liters ${CO}_2$

½ mole of $O_2$ formed 1  mole of ${CO}_2$

1 liter of $O_2$ formed 2 liters of ${CO}_2$

So, 2 liters of ${CO}_2$ will formed.

Therefore, option B is correct.