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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

Respuesta :

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
happilittlebunni

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



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