Question

Zinc metal is added to a flask containing aqueous hydrochloric acid. The flask contains 0.400 mole of HCl. How much hydrogen gas is produced if 15.5g of zinc are added to the flask?
Zn(s)+ 2HCl(aq) ⟶H2(g) + ZnCl2(aq)


1.How many moles of HCl are needed to react completely with the 15.5 g of zinc?

2. Which of the two (2) reactants is the limiting reagent?

3. How much is left over of the excessive reactant?

Answer 1

Answer:

0.4 g of hydrogen gas would be produced.

1. 0.48 mole of HCl is needed to react completely with 15.5 g of zinc.

2. HCl is the limiting reactant.

3. 2.61 g of zinc is in excess

Explanation:

From the balanced equation of reaction:

$Zn(s)+ 2HCl(aq) --> H_2(g) + ZnCl_2(aq)$

1 mole of Zn requires 2 moles of HCl to produce 1 mole of hydrogen gas.

15.5 g of zinc = 15.5/65.3 = 0.24 moles of zinc.

0.24 moles Zn is supposed to require 0.24 x 2 moles HCl which is equivalent to 0.48 moles HCl.

But only 0.400 mole of HCl is present. Hence, HCl is the limiting reagent. For complete reaction with 15.5 g of Zinc, 0.48 mole HCl would be needed.

0.400 mole of HCl will require 0.2 mole of Zn for complete reaction. This thus means that 0.24 - 02 = 0.04 mole of Zn is in excess.

0.04 mole Zn = 0.04 x 65.3 = 2.61 g excess Zn.

Now, since HCl is the limiting reagent;

2 moles of HCl is required to produce 1 mole of H2 according to the equation.

0.400 mole HCl will therefore yield 0.400 x 1/2 = 0.2 mole H2

0.2 mole H2 = 2 x 0.2 = 0.4 g H2

Hence, 0.4 g of hydrogen gas would be produced.