Question #162558
2.5 g of Al was made to react with 150 mL 0.25 M HCl. What was the limiting reactant? How many g AlCl3 was formed? How many L of H was collected if the reaction took place at 35 degree Celsius and 720 mm Hg?
Expert's answer
2Al + 6HCl = 2AlCl3 + 3H2
2.5/54= 0,0462962963
0.15 x 0.25/6 = 0,00625.
Limiting reactant is HCl.
0.00625 x 2 (27+ 3 x 35.5) = 1,66875 g of AlCl3.
0.00625 x 3 = 0,01875 moles of H2.
pV=nRT. V= nRT/p = 0,01875 x 62400 x 308 / 720 = 500,5 ml.
