THE ANSWER OF STOICHIOMETRY PROBLEMS Part 2

2. Hydrochloric acid is added to 1,3 grams of zink (Ar 65). If molarity of acid is 3 mol/L, answer the following questions.

a. 2HCl(aq) + Zn(s) --> ZnCl₂(aq) + H2(g)

2H+(aq)+2Cl-(aq)+Zn(s) --> Zn2+(aq) + 2Cl-(aq) + H2(g)

2H+(aq) + Zn(s) --> Zn2+(aq) + H2(g)

b.Determine the minimum volume of acid needed in this reaction.

n(Zn) = 1,3g/65g.mol-1 = 0,02 mol

n(HCl) = 2/1 . 0,02 = 0,04 mol

V(HCl) = 0,04 mol/3 mol/L = 0,013L= 13 mL

c. Calculate the gas produced at:

1) STP

n(H2) = 0,02 mol.
V(H2) = 0,02 mol x 22,4 L/mol= 0,448 L (448mL).

2) 1 atm and 27 degree's celcius

PV = nRT

1.V(H2) = 0,02. 0,082. 300

V(H2) = 0,492 L ( 492 mL)

3) the same T and P with nitrogen gas (Ar N = 14) that its mass of 1 L is 2 grams.

V(H2) = n(H2)/n(N2) x V(N2); n(N2) = 2/28 = 1/14 mol

V(H2) = 0,04/1/14 x 1L = 0,04 x 14 L = 0,56 L.

d. If zink used is 80% from its ore, calculate the mass of ore used.

Mass of ore = 100/80 x 1,3 g = 1,625 grams.

3. An aliage contains 90% of aluminium (Ar 27) and 10% of cuprum (Ar 63,5) is used to produce hydrogen gas. Hydrochloric acid is added to this aliage and produces 6,72 L gas at STP.

a. Write down the molecular equation and ionic equation.

2Al(s) + 6HCl(aq) --> 2AlCl3(aq) + 3H2(g)

2Al(s) + 6H+(aq)+6Cl-(aq) --> 2Al3+(aq)+6Cl-(aq)+3H2(g)

2Al(s) + 6H+(aq) --> 2Al3+(aq) + 3H2(g)

b. Calculate the mass of aliage.

n(H2) = 6,72L/22,4L.mol-1 = 0,3 mol

n(Al) = 2/3 x 0,3 mol = 0,2 mol.

m(Al) = 0,2 mol x 27 g/mol = 5,4 g

m(ore) = 100/90 x 5,4 g = 6 g.

c. If the minimum volume of acid used is 200 mL, predict the
molarity of this acid.

n(HCl) = 6/3 x 0,3 mol = 0,6 mol.

M(HCl) = 0,6 mol/0,2 L = 3 mol/L.

4. A mol of metal, M reacts with sulfuric acid 2M, produced 33,6 L hydrogen gas at STP. Answer the questions below.

a. Determine the formula of salt produced.

2M(s) + 3H2SO4(aq) --> M2(SO4)x(aq) + 3H2(g)
x = 3, Formula of salt = M2(SO4)3

n(H2) = 33,6 L / 22,4 L.mol-1 = 1,5 mol

b. Determine the oxidation number of M.

Oxidation number of M in salt is +3.

c. Write down the molecular equation and ionic equation.

2M(s) + 3H2SO4(aq) --> M2(SO4)3(aq) + 3H2(g)

2M(s)+6H+(aq)+3SO42-(aq)-->2M3+(aq)+3SO42-(aq)+3H2(g)

2M(s) + 6H+(aq) --> 2M3+(aq) + 3H2(g)

d. Calculate the minimum volume of acid needed.

n(H2SO4) = 3/3 x 1,5 mol = 1,5 mol

V(H2SO4) = 1,5 mol/ 2mol.L-1 = 0,75 L (750 mL)

5. FeS(Mr=88) is reacted with hydrochloric acid according to the following reaction:

FeS(s)+2HCl(aq) → FeCl2(aq)+H2S(g)

If this reaction produces 8 L of H2S gas, and at the same T and
P, 1 mol of H2S(g) = 20 L, answer the questions below.

a. Write down the ionic equation.

FeS(s)+2H+(aq)+2Cl-(aq) → Fe2+(aq)+2Cl-(aq)+H2S(g)

FeS(s) + 2H+(aq) → Fe2+(aq) + H2S(g)

b. Calculate the mass of FeS.

n(H2S) = 8L/20L.mol-1 = 0,4 mol

n(FeS) = 0,4 mol

m(FeS) = 0,4 mol x 88 g = 35,2 g

c. If an ore contains 70,4% FeS, calculate the mass of ore.

m(ore) = 100/70,4 x 35,2 g = 50%

d. The acid used is 4M. Determine the minimum volume of acid needed.

n(HCl) = 2/1 x 0,4 mol= 0,8 mol.

V(HCl) = 0,8 mol / 4 mol.L-1 = 0,2 L (200 mL)