Question

25. Aluminum metal shavings (10.0 g) are placed in 120.0 mL of 6.00 M hydrochloric acid. What is the maximum mass of hydrogen that can be produced (in grams)?

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

a. 11.5 g          b.   0.0505 g                         c. 4.76 g                      d. 0.727 g      

e. 31.8 g          f.   1.42 g                     g. 3.67 g                      h.   5.13 g

26. What is the maximum number of grams of ammonia, NH₃, which can be obtained from the reaction of 20.0 g of H₂ and 160.0 g of N₂

N₂ + 3H₂ → 2NH₃

a.   128 g                      b.   78.1 g        c. 21.2 g                                  d. 6.60 g       

e.   112 g                     f. 38.9 g          g. 87.0 g                                  h. 103 g

27. 4 parts, 4 points each.

A. In the following, what is the oxidizing agent? Please circle the atom (whether or not it is in a molecule.)   Do not circle anything in products, you will not get credit

Cl₂          + 2KBr      → Br₂        +       2KCl

B. What is the reducing agent in the following reaction? Please circle the atom (whether or not it is in a molecule.)   Do not circle anything in products, you will not get credit

5Fe2+(aq) + MnO4–(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l).

C. What is oxidation half reaction and reduction half reaction of Cu + Br₂ à CuBr₂? Must balance and include electrons. Electrons in each equation must match each other.

Oxidation half:

b. Reduction half:

Answer 1

We have reaction equation as :

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

moles of Al = mass of Al / molar mass of Al = 10g / 27g/mol = 0.37 mol

moles of HCl = molarity x volume in litres = 6.0 M x 0.120 L = 0.72 mol

2 mol Al reacts needs 6 mol HCl to form 3 mol H₂(g)

0.37 mol 6/2 x 0.37 = 1.11 mol

1.11 mol HCl needed but we have only 0.72 mol of HCl, and hence it is limiting reactant and the reaction will be based upon (HCl) limiting reactant

6 mol HCl forms 3 mol H₂

0.72 mol HCl forms 3/6 x 0.72 = 0.36 mol

molar mass of H₂ = 2.002 g/mol

0.36 mol = 0.36 mol x 2.002g/mol = 0.727 g

26. similarly as 25.

N₂ + 3H₂ → 2NH₃

moles of N₂ = 160g / 28g/mol = 5.71 mol

moles of H₂ = 20g / 2g/mol = 10 mol

from reaction equation we see that

1 mol N₂ needs 3 mol H₂ to form 2 mol NH₃

5.71 mol 3 x 5.71 = 17.13 mol

We need 17.13 mol H₂ , but we have only 10 mol H₂ and hence it is limiting reactant and the conversion will be based on H₂

10 mol H₂ will form 2/3 x 10 mol NH₃

= 6.66 mol

mass = moles x molar mass = 6.66 mol x 17 g/mol = 113.22 g

closest one is c = 112g

27.

Cl₂          + 2KBr      → Br₂        +       2KCl

0 -1 0 -1 oxidation no. of Cl and Br in respective molecule(not molecule)

here oxidizing agent is the one which takes electrons from other molecule and oxidize them and in due course it reduces itself by taking electrons

here Cl is oxidizing agent

similarly

5Fe²⁺(aq) + MnO₄^–(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(l)

In MnO₄^– , oxidation no. of Mn can be calculated as total charge on molecule

-1 = -2 x 4 + Mn

Mn = +7

In product it is Mn²⁺

So MnO₄^– is an oxidizing agent

C.

Cu + Br₂ ------------> CuBr₂

Oxidation half:

Cu ------------->Cu²⁺(aq) + 2e^-

Reduction half:

Br₂ + 2e^- ------>2Br^-(aq)