Zinc reacts with hydrochloric acid according to the reaction equation:
Zn(s) + 2HCl(aq) ARROW ZnCl2(aq) + H2(g)
How many milliliters of 2.50 M HCl(aq) are required to react with 8.15 g of Zn(s)?
Answer:
This is just writing the eqtn and using simple proportion maths.
Zn + 2HCl --> ZnCl2 + H2
1 mole Zn [65.4g] reacts with 2 mole HCl
65.4g Zn reacts with 2L M HCl
if HCl is 2.50M, ie 2.50times as strong. you’ll need 1/2.50of the volume
65.4g Zn reacts with 2000/2.50 mL of 2.50M HCl
8.15g Zn reacts with 2000/2.50 x 8.15/65.4 mL of 2.50M HCl
= 99.69 ml