Question

Calculate the volume of chlorine at stp that would be required to act completely with 3.70g of dry slaked line

Answer 1

Answer:

The volume required is 2.24L

Explanation:

The slaked lime Ca(OH)2, reacts with chlorine, Cl2, as follows:

6Cl₂(g) + 3Ca(OH)₂(s) → Ca(ClO₃)₂ (aq) + 2CaCl₂ (aq) + 6HCl (aq)

Where 6 moles of chlorine react with 3 moles of slaked llime,

To solve this question we must find the moles of slaked lime added. With these moles we can find the moles of chlorine required and its volume at STP as follows:

Moles Ca(OH)2 - Molar mass: 74.093g/mol-

3.70g * (1mol / 74.093g) = 0.0500 moles Ca(OH)2

Moles Cl₂:

0.0500 moles Ca(OH)2 * (6 mol Cl₂ / 3 mol Ca(OH)2) =

0.100 moles Cl₂

Now using PV = nRT

nRT / P = V

Where n are moles: 0.100 moles

R is gas constant = 0.082atmL/molK

T is absolute temperature = 273K at STP

P is pressure = 1atm at STP

And V is volume, our incognite:

0.100mol*0.082atmL/molK*273K / 1atm = V

2.24L = Volume of Cl₂

The volume required is 2.24L