Question

 

What theoretical yield (grams) of hydrogen can be obtained from 0.0334 Mg and 15.00mL of 6.0 M HCL?

Answer 1

Mg(s) + 2HCl----------> MgCl2 + H2

no of moles of Mg = W/G.A.Wt

                               = 0.0334/24 = 0.00139moles

no of moles of HCL   = molarity * volume in L

                                  = 6*0.015 = 0.09 moles

2 moles of HCL react with 1 moles of Mg

0.09 moles of HCl react with = 1*0.09/2   = 0.045 moles of Mg

   Mg is limiting reactant

1 mole of Mg react with HCl to gives 1 mole of H2

0.00139 moles of Mg react with HCl to gives = 1*0.00139/1   = 0.00139 moles of H2

mass of H2 = no of moles * gram molar mass

                       = 0.00139*2   = 0.00278g