Calculate the volume of the chlorine that would be required to react completely with 3.70g of dry slaked lime according to the equation Ca (OH) 2+Cl=CaOCl2+H2O

H = 1, O=16, CA=40,1 mole of gas occupies 22.4dm3 at s. t. p

1.12 dm³

Explanation:

Given dа ta:

Volume of chlorine = ?

Mass of slaked lime = 3.70 g

Solution:

Chemical equation;

Ca (OH) ₂ + Cl₂ → CaOCl₂ + H₂O

Number of moles of slaked lime:

Number of moles = mass / molar mass

Number of moles = 3.70 g / 74.1 g/mol

Number of moles = 0.05 mol

Now we will compare the moles of Ca (OH) ₂ and chlorine.

Ca (OH) ₂ : Cl₂

1 : 1

0.05; 0.05

Volume of chlorine:

volume of one mole of gas at stp = 22.4 dm³

0.05 * 22.4 dm³ = 1.12 dm³