We have reaction equation as : 2Al(s) + 6HCl(aq) 2AlCl 3 (aq) + 3H 2 (g) moles o
ID: 699845 • Letter: W
Question
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)
Explanation / Answer
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:
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).
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