Question

In the reaction, Zn(s) + 2 HCl(aq) --> ZnCl2 (aq)+ H2(g), 25 grams of Zn are reacted with 17.5 g of HCl. How many grams of H2 will be produced?

Answer 1

Answer:

0.480 g of H₂ are produced, in the reaction.

Explanation:

This is the reaction:

Zn(s)  +  2HCl → ZnCl₂ (aq) +  H₂ (g)

We havethe mass of both reactants, so we must work with them to find out the limiting reactant and then, determine the amount of H₂ produced.

Let's convert the mass to moles ( mass / molar mass)

25 g / 65.41 g/mol = 0.382 moles Zn

17.5 g / 36.45 g/mol = 0.480 moles HCl

Ratio is 1:2, so 1 mol of Zn react with the double of moles of HCl.

0.382 moles of Zn would need the double of moles to react, so (0.382 .2) = 0.764 moles of HCl. → We only have 0.480 moles, so the acid is the limiting.

Now let's determine the moles of H₂ formed.

Ratio is 2:1, so If i take account the moles I have, I will produce the half of moles of my product.

0.480 moles / 2 = 0.240 moles of H₂ are produced.

To find out the mass, we must multiply mol . molar mass

0.240 mol . 2g/mol = 0.480 g