Answer to Question #339119 in General Chemistry for Tango

Question

A mass of 200 g of Zn reacts with 800.0 mL of 4.00 M HCl to form Zn(Cl)₂

Zn(s) + HCl (aq) → Zn(Cl)₂ (aq) + H₂ (g)

1.1      Balance the equation of the chemical reaction                                                                                   

1.2      Identify the liming reactant.                                                                                                                                  

1.3      Determine the excess reactant remaining after the reaction was complete.                                      

1.4      Calculate the theoretical yield.                                                                                                                             

1.5      Calculate the percentage yield for the production of H2. The actual yield obtained for H₂ was 1.46 g.

Answer 1

With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:

2 HCl (aq) + Zn (s) → ZnCl₂ (aq) + H₂ (g)

A sample of pure zinc with a mass of 5.98 g is reacted with excess hydrochloric acid and the (dry) hydrogen gas is collected at 25.0 ˚C and 742 mm Hg. What volume of hydrogen gas would be produced?

Solution

This is a “single state” problem, so we can solve it using the ideal gas law, PV = nRT. In order to find the volume of hydrogen gas (V), we need to know the number of moles of hydrogen that will be produced by the reaction. Our stoichiometry is simply one mole of hydrogen per mole of zinc, so we need to know the number of moles of zinc that are present in 5.98 grams of zinc metal. The temperature is given in centigrade, so we need to convert into Kelvin, and we also need to convert mm Hg into atm.

Conversions:

25.0C+273=298K

25.0C+273=298K

(742mmHg)×(1atm760mmHg)=0.976atm

(9.6.2)(742mmHg)×(1atm760mmHg)=0.976atm

(5.98gZn)×(1.00mol

65.39gZn)=0.0915mol

(9.6.3)(5.98gZn)×(1.00mol65.39gZn)=0.0915mol

Substituting:

PV=nRT

(9.6.4)

(9.6.4)PV=nRT

(0.976atm)×V=(0.0915mol)(0.0821Latmmol−1K−1)(298K)(9.6.5)(0.976atm)×V=(0.0915mol)(0.0821Latmmol−1K−1)(298K)

V=(0.0915mol)(0.0821Latmmol−1K−1)(298K)(0.976atm)

=2.29L