Question

What volume of H2(g) is produced when 3.20 g of Al(s) reacts at STP?

Answer 1

H2(g) is produced when Al reacts with HCl.

The balanced equation between HCl and Al(s) is

2Al + 6 HCl ------------> 2AlCl3 + 3H2

2 mol 3 mol

2 x 27 g = 54 g 3 mol [ molar mass of Al = 27 g]

3.2 g ?  

Then,

? = ( 3.2 g/ 54 g) x 3 mol of Al

= 0.178 mol of Al

Therefore, 0.178 moles of Al will be formed.

Now we have to calculate the volume of Al at STP.

At STP, T = 273 K , P = 1 atm

Ideal gas equation is

PV = nRT

V = nRT/P

= 0.178 mol x x 0.0821 L.atm/mol/K x 273 K / 1 atm

= 3.98 L

Volume of H2 = 3.98 L

Therefore,

volume of H2(g) is produced when 3.20 g of Al(s) reacts at STP = 3.98 L