Question

Zinc reacts with hydrochloric acid according to the reaction equation

How many milliliters of 4.00 M HCl(aq) are required to react with 7.95 g of Zn(s)?

Answer 1

Zn + 2HCl --> ZnCl2 + H2

mass of Zn = 7.95 g
molar mass of Zn =65.38 u
so moles of Zn present = mass/ molar mass = 7.95 g/65.38 u =
0.1215968 mols

According to balanced equation 2 mole HCl needed for 1 mole Zn

SO mole rato = 1:2

moles of HCl needed = 2*0.1215968 mols = 0.2431936 mol

given molarity of HCL = 4.00 M
volume of HCl = molarity *number of mols = 0.2431936 mol* 4.00 M
= 0.97277 Liters

volume of HCl in ml = 972.77 ml
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