Question

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​

Answer 1

Answer:

1.12 dm³

Explanation:

Given data:

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