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
