If a sample of 27.0 g of aluminum metal is added to 333 mL of 3.0 M HCL acid, the volume of hydrogen gas produced at standard temperature and pressure is

Question

a. 2.8 L
b. 5.60 L
c. 11.2 L
d. 22.4 L

Answer 1

6HCl + 2Al>>2AlCl3 + 3H2
moles aluminum=27/27=1mole which means you need 3 moles HCL.
moles HCl=.333*3=1.0 moles
so the limiting reactant is HCL, giventhat, moles H2 generated=1/6 * 3*22.4 liters.

Answer 2

a,b