Question

How many grams of Al would be required to form 245 g AlCl₃?

How many molecules of Al are required to fully react with 100.0 g Cl₂?

12.5 mL of 12.0 M HCl (aq) is transferred to a flask and an additional 200.0 mL of water is added. Determine the molar concentration of the new, dilute HCl (aq).

Answer 1

rower: moles of Aklz = 245 133-34 = 1.837 & mole of all contains a mole of Al. 1.837 moles of All will contain 1.837 moles of me mass of Al = 1.837 x 26.98 = 49.579 =49.599 AO + 3 ch - All moles of d = 100 = 1.41 from stoichemitry & moles of q require 70.91 1 mole of ne 80 loul " " " will rezecure = 1 x 1.41 = qogu mole of Al.

molecules of Al = 0.94 x 6.022 x 10²3 = 5.66x 1023 Nv. = N , y final volume of nel Jautial Come of not 6 zatial volume of not final conco of ne 12 x 12.5 = *21255 M = 0.705 M The conch of dil, hee is 0.705M.