Question

III(15pts) Some commercial drain cleaners make use of the following reaction to declog drains. NaOH(aq) + 2 Al(s) +6 H (1) ...> 2 NaAl(OH)/(aq) + 3 H2(g) 18 4,63 grams of Al(s) are reacted with excess NaOH and water, what volume of H.(R) can be produced at 742 mmHg, and 23.8 "C?

Answer 1

Given neaction NaOH (aq) + 241/3) + 616010) - 2 Na Al(OH)4 (aq) + 3tl(g). from the above reaction we obsenned that from 2 mole of Al(s) we gain 3 mole of Hy (9) AICS) is the limiting nogen as other sogante are present in excess. Amount of Alli) - 4139 Molan mass of Al : 27 gmott So , number of moles of Al .. Mass Molan mas 4.63g 27 amal 3 0'171 money 2 mole of Alls) gives 3 mole of tz (4) So, 011 mole of Alo) gines x 0111 mol of lly 2

- 0256 moles of Hly (g).(m) n - nomber of moles of lg. Also given Pressure, P: 742 mm Hg = 0·00131579 y 742 afrom = 0.976 afm (1 multig = 0·00131519 aton) Temperature, T = 23.8°C = (273 + 23.8) K - 296.3K using ideal gas equation » PV = nRT R = 0.0820 Lafm markt (has constant Now A Jo 0.976 atm x V - 0256 molx 0082 Latm moltki x 2903K

- C atm mollk" 0.256 mol x 0.082 x 296.3K 0'976 alm = 6.37 L So, volume of H₂ (g) is 6:37