Question

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? 0.375 M 1.50 M 3.00 M 6.00 M

Answer 1

Answer: D on Edge :)

Step-by-step explanation:

took the test

Answer 2

`\boxed{\text{6.00 mol/L}}`

Step-by-step explanation:

(a) Balanced equation

2HCl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O

(b) Moles of Ca(OH)₂

`\text{Moles of base} = \text{2.00L} * \frac{\text{1.50 mol}}{\text{1 L}} = \text{3.000 mol base}`

(c) Moles of HCl

`\text{Moles of HCl} = \text{3.000 mol base} * \frac{ \text{2 mol HCl}}{\text{1 mol base}} = \text{6.000 mol HCl}`

(d) Molar concentration of HCl

`\text{Molar concentration} = \frac{\text{moles of solute}}{\text{litres of solution}}\\\\c = ( n )/( V)\\\\c= \frac{ \text{6.000 mol}}{ \text{1.000 L}} = \text{6.00 mol/L}`

The molar concentration of the HCl was
`\boxed{\textbf{6.00 mol/L}}`