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A 1.00 L volume of HCl reacted completely with 2.00 L of 1.50 M Ca(OH)2 according to the balanced chemical equation below. 2HCl + Ca(OH)2 CaCl2 + 2H2O What was the molarity of the HCl solution? A)0.375 M B)1.50 M C)3.00 M D)6.00 M

Respuesta :

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

n(HCl)=c₁v₁
n{Ca(OH)₂}=c₂v₂
n(HCl)=2n{Ca(OH)₂}

c₁v₁=2c₂v₂

c₁=2c₂v₂/v₁

c₁=2*1.50*2.00/1.00=6.00 mol/L

D) 6.00 M HCl

Ans: D) 6.00 M

Given:

Volume of HCl, V(HCl) = 1.00 L

Volume of Ca(OH)2, V(Ca(OH)2) = 2.00 L

Molarity of Ca(OH)2, M(Ca(OH)2) = 1.50 M

To determine:

Molarity of HCl, M(HCl)

Explanation:

The reaction is:

2HCl + Ca(OH)2 → CaCl2 + 2H2O

[tex]Moles Ca(OH)2 = M(Ca(OH)2) * V(Ca(OH)2)\\\\= 1.50 moles/L * 2.00 L = 3.00 moles[/tex]

Based on the reaction stoichiometry:

2 moles of HCl reacts with 1 mole Ca(OH)2

Therefore, 3.00 moles of Ca(OH)2 would react with

= [tex]\frac{3.00 mole Ca(OH)2*2 moles HCl}{1 mole Ca(OH)2}  \\\\= 6.00 moles HCl[/tex]

Given that V(HCl) = 1.00 L

[tex]Molarity = \frac{moles HCl}{Volume HCl} \\\\= \frac{6.00 moles}{1.00 L} = 6.00 M[/tex]

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