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)
<em>Where 6 moles of chlorine react with 3 moles of slaked llime,</em>
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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:
<em>Moles Ca(OH)2 - Molar mass: 74.093g/mol-</em>
3.70g * (1mol / 74.093g) = 0.0500 moles Ca(OH)2
<em>Moles Cl₂:</em>
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
<em>Where n are moles: 0.100 moles</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is absolute temperature = 273K at STP</em>
<em>P is pressure = 1atm at STP</em>
<em>And V is volume, our incognite:</em>
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0.100mol*0.082atmL/molK*273K / 1atm = V
2.24L = Volume of Cl₂
<h3>The volume required is 2.24L</h3>
