How many milliliters of 3.00 M HCl(aq) are required to react with 6.35 g of an ore containing 28.0% Zn(s) by mass?
Zn (s) + 2 HCl (aq) -------> ZnCl2 (aq) + H2 (g)
mass of Zn = (6.35 *28 /100) = 1.778 g of Zn
number of moles of Zn = 1.778/ 65.39 = 0.02719 moles
1 mole of Zn requires 2 moles of HCl
number of moles of HCl = 0.02719*2 = 0.05438 moles
volume of HCl = (0.05438 / 3) L
= 18.13 mL of 3 M HCl is required
How many milliliters of 3.00 M HCl(aq) are required to react with 6.35 g of an...
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