Question

Calculate the volume of H2(g) at 273 K and 2.00 atm that will be formed when 275 mL of 0.725 M HCl solution reacts with 50.0 g Zn(s) to give hydrogen gas and aqueous zinc chloride. (R = 0.08206 L atm/K mol)

Answer 1

Answer:

The volume of  H₂ (g) obtained is 22.4L

Explanation:

First of all, think the reaction:

2HCl (aq) + Zn (s) → ZnCl₂ (aq)  + H₂ (g)

You have to add a 2, in the HCl to get ballanced.

Now we should know how many moles of each reactant, do we have.

Volume . Molarity = moles

Notice that volume is in mL, so I must convert to L.

275 mL = 0.275L

0.275L . 0.725mol/L = 0.2 moles of HCl

Molar mass of Zn: 65.41 g/m

50 g / 65.41 g/m = 0.764 moles

Ratio between reactants is 2:1, so I need the double of moles of HCl to react, and a half moles of Zn to react.

My limiting reactant is the HCl, for 0.764 moles of Zinc, I need 1.528 (0.764 .2) of HCl, and I only have 0.2 moles.

Ratio between HCl and H₂ is 1:1, so 0.2 HCl make 0.2 moles of gas

Now apply the Ideal Gas Law, to find out the volume

P. V = n . R . T

2 atm . V = 0.2 mol . 0.08206L atm/K mol . 273K

V =  (0.2 mol . 0.08206L atm/K mol . 273K ) / 2 atm

V = 2.24 L