How many ml of 1.257M HCl are needed to completely react with 32.75g of Mg metal?

Question

How many ml of 1.257M HCl are needed to completely react with 32.75g of Mg metal?

Answer 1

First the balanced reaction to get the mole ratio: Mg + 2 HCl --> MgCL₂ + H₂. Then stoichiometry to find moles of HCl: (32.75g Mg) * (1 mole Mg /24.305g Mg) * (2 mole HCl /1 mole Mg) = 2.695 mole HCl.

Now let x be the desired volume in mL: (M = 1.257 mole/L) * (x/1000 L) = 2.695 mole, x/1000 = 2.695/1.257 = 2.144 L, x = 2,144 mL.