Question

1)Use: 2Al + 6HCl → 3H2 + 2AlCl3

a) If 5.34 g H2 are formed, how many grams of Al reacted?

b) If 2.500 kg Al reacts, how many moles of H2 are produced?

c) If 6.13 moles of Al react, how many moles of HCl need to react?

d) If 125.00 moles HCl reacts, how many mg of AlCl3 are produced? e) If 4.81 x 1024 formula units of AlCl3 are produced, how many moles of HCl reacted?

Answer 1

1.a Al reacted =47.65g

b. H2 produced = 280.19 g

c. HCl reacted = 18.39 mol

d. AlCl3 produced = 5.556 × 10⁶ mg

e.moles of HCl reacted = 24.116 mol

Page NO. 2 Al + 6Hce 3H + 2 - 2 Alce Alce 5.34 g of 4 are Joomed No. of moles of H = 5,34 g - 2.65 mol good 3 mol H₂ are formed by a mol Al b a 2.65 mol H 2 2.65 mol Al 2 3 = 1.766 mol se amount of Al Reacted = 1,966 x 26.981539 g 8 en 47.66g b) 2.5 kg Al Reacts a la 99 66 - No.of moles of Al = 2.5&log 0 26.981539 mot 2 mol Al produce 3 mol H 92.66 mol al = 2 x 22.66 mol H₂ = 138.99 mol H₂ Amount of He produced = 138.99 x 2.0169 30 og = 280.194 2 c) 6.13 mol Al weats 2 mol Al racts with 6mol He 1 6.13 mol A = 6 x 6.13 mol hee = 18.39 mol Ha. d. 125 mol HCl reacts . 6 mol Hee produce a mol Alle 125 mol Hel 2x 125 m d dece 6 z 41.667 mol Alce

Orarlie amount of A Date - Page No. 3 400 led led = 41.667 x 133.340549 =5555,8564 E 5,556 x 100 mg . 4.81 % 10 24 Jamula units of alll are produced A 1 mol contornes 6.022 x 10 23 0 formulo units! I No. of moles of Allo b = 4,812 24 6 smase 6020 x 1023 H - = 0.80312 xlor lb hom aa8.0312 mol | 2nd Alces are produced by bind Heel 8.0312 md Alle e 8 0312 mol Hee - = 24. 1116 md he ta .e asil P - olxa. - BA od 0 3 D 2 a