7]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? ca (s) + 2hcl (aq) yields cacl2 (aq) + h2 (g) 1200 ml 2400 ml 3750 ml 9600 ml

I used stoichiometry here:

60 g Ca * 1 mol Ca/40.078 g Ca * 2 mol HCL / 1 mol Ca * 1 L HCL Solution/1.25 mol HCl * 1000 ml HCL solution / 1 L HCL solution = 2395.3291 mL

So, 2400 mL is your answer

According to the chemical equation, the ratio between Ca and HCl for a complete reaction is 1:2. The moles of Ca can be calculated by mass/molar mass, thus 60.0 g/40 g/mole = 1.5 mole. So the moles of HCl in the solution is 1.5mole*2 = 3 mole.

So the volume of HCl solution = Moles of HCl / concentration of HCl = 3 mole / 1.25 M = 2.4 L = 2400 mL