Question

Fe(s) + 2HCl(aq) --> FeCl2(aq) + H2(g)

When a student adds 30.0 mL of 1.00 M HCl to 0.56 g of powdered Fe, a reaction occurs according to the equation above. When the reaction is complete at 273 K and 1.0 atm, which of the following is true?

A) HCl is in excess, and 0.100 mol of HCl remains unreacted.

D) 0.22 L of H2 has been produced.

The correct answer is D. I can't figure out why A is wrong.

Answer 1

30.0 mL(= 0.030L)   of 1.00 M HCl = 0.030 L x 1 M = 0.03 mole

0.56 g of powdered Fe = 0.56 gm / (56 gm /mol ) = 0.01 mole (where atomic mass of Fe =56 gm /mol)

Fe(s) + 2HCl(aq) --> FeCl2(aq) + H2(g)

1 mole 2mole 1 mole 1 mole

1 mole of Fe reacts with 2 moles of HCl

or, 0.01 mole of Fe reacts with (2 x 0.01 ) = 0.02 moles of HCl

available HCl = 0.03 mole

Thus, HCl that remains unreacted = 0.03 -0.01= 0.02 mole ( Thus option A is incorrect)

Fe is the limiting reagent.

i mole Fe gives 1 mole hydrogen gas

or, 0.01 mole Fe gives 0.01 mole hydrogen gas

volume of Hydrogen gas, V = nRT / P (Ideal gas law)

= (0.01 mole x 0.082 L.atm/mol.K x 273K) / 1 atm

= 0.22 L