Question

A 0.2100-g sample of an Al-Zn alloy reacts with HCl to form hydrogen gas: 2 Al(s) + 6 H+ (aq) → 2 A1+ (aq) + 3 H2(g) Zn(s) + 2 H+ (aq) → Zn2+ (aq) + H2 (8) The hydrogen produced has a volume of 0.117 L at 24°C and 755 mm Hg. What is the percentage of zine in the alloy? Percentage of zinc = Submit Answer Try Another Version 8 item attempts remaining

Answer 1

traction let us o of zinc in allay be a 2A +6H - 2A(²+ + 3H2 From reaction of we can say Imole of Ac produces 3 mole of H₂ 2 19 of AL produces = (3x2 g of H₂ = ( ) g of Hz Zn + 2H+ 2x2 + H₂ I bj treh - From reaction Imole zinc produces ImoleqHz (100 lg of an produces 2 / 9 H 2 O >> Igneel of 2n-Al alioy will produce - Hor a su + gieten) (1-) Temperatwee 724°C = 24 +273 = 297 K Pressure = 755mm = 755 = 0.9934 alm. Volume of hydrogen = 0.117L By ideal gas equation, opvanRTs 0.9934x0.117 = n*0.082* 297 ha0.9934x0.117 0.082 X 297 = 4.77 x 10-3 male 760

Weight of H₂ Produced = 4.77x10²3x2 =9.54x1039 So, Ig of zu-Al produce = 2. 0219 of 2n-Al produce = 0.21 (34tý (1-) 9.5481013 = 0.21 (žu+*(1-1) ) 0.0454 = 0.03074 +0.11 -oeltu Dung u= 0:1l-00454 0.11 -0.0307 = 0.8146. ADP Lab % of Zn=0.81 46X100 LPSXS 80 foy781. 46 y.