68.7mL
Step-by-step explanation:
• 1st. we write the reaction
2HCI(ag) + Ca(OH)2(s)
› CaCI2 (ag) + 2H20
(1).
2ns, we find moles of Ca(OH)2
moles= concentration x volume = 0.125mol/L
X0.05L=6.25X10^-3moles
• we find moles of HCI
mole ratio between HCI and Ca(OH)2 is 2:1
thus moles of HCI=2/1 x
6.25X10^-3moles=1.25x10^-2moles
• we find volume of HC
the concentration is 0.182mol/L
we know that 1L=0.182mol
thus =1.25X10^-2moles / 0.182mol x
1L=0.0687L
Volume of HCI required is 68.7mL