Question

# Zinc reacts with hydrochloric acid according to the reaction equation Zn(s) + 2HCl(aq) -> ZnCl2(aq) +...

Zinc reacts with hydrochloric acid according to the reaction equation

Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)

How many milliliters of 6.50 M HCl(aq) are required to react with 3.05 g of Zn(s)?

#### Homework Answers

Answer #1

Number of moles of Zn = 3.05 g / 65.38 g/mol = 0.0467 mole

from the balanced equation we can say that

1 mole of Zn requires 2 mole of HCl so

0.0467 mole of Zn will require

= 0.0467 mole of Zn *(2 mole of HCl /1 mole of Zn)

= 0.0934 mole of HCl

molarity of HCl = number of moles of HCl / volume of solution in L

6.50 = 0.0934 / volume of solution in L

volume of solution in L = 0.0934 / 6.50 = 0.0144 L

1L = 1000 mL

0.0144 L = 14.4 mL

Therefore, the volume of solutiion is 14.4 mL

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