Zinc reacts with hydrochloric acid according to the reaction equation Zn (s)+2HCl (aq)--> ZnCl2 (aq) + H2(g) How many milliliters of 2.50 M HCl(aq) are required to react with 3.55 g of an ore containing 37.0% Zn(s) by mass?
Zn (s) + 2HCl (aq)--> ZnCl2 (aq) + H2(g)
1 mole 2 moles
3.55gm of ore containing 37%Zn
mass of Zn = 3.55*37/100 =1.3135 gm
no of moles of Zn = 1.3135/65.4 = 0.02 moles
From the balanced equation
1 mole of Zn react with 2 moles of HCl
0.02 moles of Zn react with = 2*0.02/1 = 0.04 moles of HCl
Molarity = no of moles /volume in L
2.5 = 0.04/volume in L
volume in L = 0.04/2.5 = 0.016 L
volume in milliliters = 0.016*1000 = 16ml
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