Question

Zinc and sodium metal each react with sulfuric acid according to the following equations. Zn(s) + H2SO (aq) + ZnSO4 (aq) + H (8) 2 Na(s) + H, SO, (aq) Na, 50, (aq) + H,() A 26.06 g mixture of zinc and sodium is reacted with a stoichiometric amount of sulfuric acid. The reaction mixture is then reacted with 144 mL of 3.44 M barium chloride to produce the maximum possible amount of barium sulfate. Determine the percent sodium by mass in the original mixture. mass percent Na : If 82.05 mL of sulfuric acid was added, what was the concentration of the sulfuric acid? concentration:

Answer 1

(Let us assume x grams be the weight of Zn, Then weight of Na is (26.06-x) grams)

Zn(s) + H2SO409) + ZnSO40g) + H26
;
V 99 = 'osuzu "99 = 'oshu 99 = uzu il

2 Na(s) + H250409) + Na2SO40g) + H26
;
26.06 - 2 26.06 - I 22 26.06 – n Na = TH2SO4 = nNa2SO4 = 23 x 2: Na2SO4 = - 23 x 2

On adding BaCl2 the following reactions occur

ZnSO409) + BaCl2aa + BaSO4 + ZnCl2
;
Na2SO409) + BaCl2 + BaSO4 +2 NaCl(aq)

M = n x 1000 - - 3.44 x 144 in BaCl = - -= 0.49536 1000 VmL

n ZnSO4 + n Na2SO4 = n Bach
;
2 26.06 - 23 x 2 3.44 x 144 1000

1 = 11.6349 grams

Concentration of H2SO4 can be calculated as :

WtNa= 14.965 grams mass percent of Na= 57.4253

MH so, = "H2S04 X 1000 Mw.so = nH2SO4 x 1000 Vm 82.05

TH2SO = n BaCl2
  
:: MH2SO4 = 0.49536 x 1000 -= 6.0373 M 82,05