How many milliliters of a 1.25 molar hydrochloric acid (HCl) solution would be needed to react completely with 60.0 grams of calcium metal? (2 points)

Ca (s) + 2HCl (aq)yields CaCl2 (aq) + H2 (g)

Respuesta :

Answer:

2400 mL

Explanation:

[tex]Ca + 2HCl \implies CaCl_2 + H_2[/tex]

According to this equation, the stoichiometric ratio between [tex]Ca[/tex] and [tex]HCl[/tex] for the complete reaction is 1:2.

We know that the number of moles of [tex]Ca[/tex] can be calculated using the mole formula. (number of moles = mass / molar mass)

Moles of Calcium = [tex]\frac{60}{40}[/tex] = 1.5 mol

So the moles of [tex]HCl[/tex] = [tex]1.5 \times 2[/tex] = 3.0 mol

Volume of HCl solution = Moles of HCl/ concentration of HCl

Volume of HCl solution = [tex]\frac{3}{1.25}[/tex] = 2400 mL