Question

Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)

How many liters of H2 would be formed at 680 mm Hg and 17 ∘C if 27.0 g of zinc was allowed to react?

How many grams of zinc would you start with if you wanted to prepare 6.00 L of H2 at 400 mm Hg and 35.0 ∘C?

Answer 1

From the equation of the reaction, we see that the number of moles of H₂ formed is equal to the number of moles of Zn that reacts. Therefore, we find the number of moles of Zn.

Number of moles = mass / molar mass = 27 / 65.98 = 0.41 moles

Number of moles of H₂ = 0.41 moles

From the ideal gas equation, we have: V = n * R * T / P

Using the values in the question, converting Pressure to atm and temperature to Kelvin, we have:

V = 0.41 * 0.0821 * 290 / [680/760]

V = 10.91 L

The next part of the problem is simply the reverse of what we did here, i.e finding the mass of Zn from Volume of H₂. We first find the number of moles of H₂

n = P * V / [R *T]

n = (400/760) * 6 / [0.0821 * 308]

n = 0.125 moles

This is equal to the number of moles of Zn that reacts.

Mass of Zn = number of moles * molar mass = 0.125 * 65.98 = 8.25 g