Answer:
197 ml of 4.0M HCl
Explanation:
What is the minimum volume of 4.00 M HCl (aq), in mL, needed to completely react with 7.10 g of aluminum?
first write and balance the equation
2Al + 6HCl-----------> 2AlCl3 + 3H2
so
for every 2 mol of Al, we need 6 mol of HClf
so for every mol of Al we need 3 mol HCl
Al has an atomic mass of 27 gm/mol
so
7.10 gm Al is 7.10 gm// (27 gm/mol) = (7.10/27 ) mol =0.263 mol
so
we need 0.263 X 3 =0.789 moles of Cl
the HCl solution is 4.00M or 4 moles/ 1000ml
we need y ml of this 4.00M HCl to give us 0.789 moles of HCl
so
(0.789/Y =4.00
0.789 = 4.00Y
Y = 0.789/4.00 = 0.197 liters =
197 ml
check 197 ml of 4.00M HCl will have
0.197 X 4.00 moles = 0.788 moles HCl
at a 3:1 mol ratio with Al this will react with 0.788/3 moles of Al.
=0.263 moles
atomic mass of Al is 27.gm/mol
0.263 mols of Al weigh 0.263X27 = 7,10 gm Al which is what we habe
