What volume of h2(g) at 27oc and 680 torr is produced by the reaction of 15.0 g al metal with excess hcl(aq) to produce alcl3 and h2 (hint: you may want to write and balance an equation first)?

Step 1 :

The balanced chemical equation for the above problem can be written as:

2Al + 6HCl → 2AlCl₃ + 3H₂

Step 2 :

Since 15 grams of Al metal react with the excess of HCl, therefore Al will be the limiting reactant.

The moles of Al that reacted will be = 15/27

=0.55 moles of Al.

Step 3:

Molar ratio of Al and H₂ will be 2:3.

Step 4:

Therefore moiles of H2 produced by 0.55 moles of Al

0.55 moles of Al × (3 moles of H₂ ÷2 moles of Al)

=0.55×1.5

=0.825 moles of H₂.

Step 5:

By using the ideal gas equation we can obtain the volume of H2 produced.

pV=nRT (Ideal Gas Equation)

V=(nRT) ÷ p

Step 6:

n=0.825 moles

R=0.0821 litre atm mol⁻¹ K⁻¹

T=27 degree celsius = 27+273 K= 300 K.

p = 680 torr × ( 1 atm ÷ 760 torr )

=0.894 atm.

Step 7:

V=[0.0825 moles × 0.0821 atm L mol⁻¹ K⁻¹×300 K] ÷ 0.894 atm

V=2.27×10⁻³ litres.

i'm assuming that the correct answer is indeed the first one/ "a"..