Question

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

Part A: How many liters of H2 would be formed at 538 mm Hg and 16∘C if 22.0 g of zinc was allowed to react?

Part B: How many grams of zinc would you start with if you wanted to prepare 5.65 L of H2 at 350 mm Hg and 34.5 ∘C?

Answer 1

- 65.38 Ans. Given reaction - Zn (s) + & FC (aq? Zn C (4) + 1, q2 Atomic may og zn = 65.38gmoi! de toit Part A Calculation of volume of Me - first calculate the the moles of He produced from the above eaction 65.38g oy zn produces dg Me 22 g of Zn will produce Hal = 2 x 22 g of Ha = 0.67 g of Ha Mole of He produced = Mass of H2 -0.679 Molar mass of Hq - agmor! This = 0.35 mol [PV = nRT I P = 538mmHg, T= 16°C = 289 k (16t273) i atm = 760 mm Hg 760 mm Hg = lam 538 mm rig = L x 538 atm 0.335 mol 760 = 0.7078 atm R= 0.0821 L arm ki molt

VHg = 0.335 mol X 0.0821 L atm k'mol "x 289 k 0.7078 alm Villa Ellaan Volume of my formed is 11.229 L Part B. Himount of Zina = ? Calculate the no. of moles of He produced PV = nRT P= 350 mmHg P = 350 atm = 0.460 atm 760 T = 34.5°C - 34.5+273 = 307.5 K V= 5.65 L Niue = b = 0.460 atm x 5.654 0.0821 Lamm K Imoi'x 307.5k 25.24575 Mhly = 4.599. mol = 0.103 mol 1778 - 0-103 mol Amount of the produced Cing) = moles x molar May Che) = 0.103 mol x & g mort = 0.2069 2 g of Hy produced from 65.389 of an c from the Reaction) 0.206 g of His will produce In = 65380.206 g of an = 6.734 g of an I Amount of zinc = 6.734 9