Calculate the amount of H2 produced from the reaction of 4.25 grams Zn and 25 ml of 3.0 M HCL.
Zn + 2HCl => ZnCl2 + H2
Moles Zn = (4.25g/65.37)= 0.065mole Zn available.
Moles HCl = 25ml(3M HCl) = 0.025(3) mole HCl = 0.075 mole HCl
HCl is limiting reagent => Only 1/2(0.075 mole)Zn = 0.0375 mole Zn will be consumed in reaction => (0.065 - 0.0375)= 0.0275 mole Zn in excess.
From balanced equation, moles Zn used = Moles H2 produced = 0.0375 mole H2 => 0.0375(2)gram H2 = 0.075g H2 produced.
If Volume is needed, assume STP conditions => Vol H2 = nRT/P = (0.0375mole)(0.08206L-atm/mole-K)(273K)/(1atm) = 0.840 Liters H2 at STP or, use Vol H2(STP) = 22.4L/mole(0.0375mole) = 0.840 Liter H2.