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## Chemical Reactions Exam2 and Problem Solutions

Chemical Reactions Exam2 and  Problem Solutions

1. In which one of the following compounds, X has different oxidation state?

I. HXO4-

II. X2O7-2

III. XO4-2

IV. XO3

V. XO2

Solution:

Oxidation states of H is +1 and O is -2. Using them we find oxidation state of X in given compounds.

I. In HXO4- compound, sum of oxidation states is -1. So,

+1+X+4.(-2)=-1

X=+6 (Or X=-2)

II. In X2O7-2 compound, sum of oxidation states is -2. So,

2X + 7.(-2)=-2

X=+6 (Or X=-2)

III. In XO4-2 compound, sum of oxidation states is -2. So,

X + 4.(-2)=-2

X=+6 (Or X=-2)

IV. In XO3 compound, sum of oxidation states is 0. So,

X + 3.(-2)=0

X=+6 (Or X=-2)

V. In XO2 compound, sum of oxidation states is 0. So,

X + 2.(-2)=0

X=+4 V is different from others.

2. In which one of the following reactions, N is oxidized?

I. 3Cu + 2NO3- + 8H+ → 3Cu+2 + 2NO + 4H2O

II. Cl2O + 4NO2 + 3H2O → 2Cl-+ 4NO3- + 6H+

III. 2Ag+ + 2NH3 + H2O2 → 2Ag + 2NH4+ + O2

IV. 2N2O5 → 4NO2 + O2

V. 2NO2 → 2NO + O2

Solution:

Elements which give electrons are reduced and which gain electrons are oxidized. Thus, we must find choice in which N lose electron.

I. NO3- → NO We find oxidation state of N in both sides. (Oxidation state of O is -2)

x + 3(-2) = -1, x=+5 in reactants

x + (-2) = 0, x=+2 So, N is reduced

II. NO2 → NO3-

x + 2(-2) = 0, x=+4 in reactants

x + 3(-2) = -1, x=+5 So, N is oxidized

III. NH3 → NH4+1

x + 3(1)=0, x=-3 in reactants

x + 4(1)=+1, x=-3 so oxidation state of N does not change

IV. N2O5 → NO2

2x + 5.(-2)=0, x=+5 in reactants

x + 2.(-2) =0, x=+4, So, N is reduced

V. NO2 → NO

x + 2(-2)=0, x=+4 in reactants

x + (-2)=0, x=+2 Thus, N is reduced.

3. Using 40 g Fe2O3 and 1,5mol CO following reactşons occur in closed container with 100% efficiency.

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

Which ones of the following statements are true for this reactşon?(Fe=56, O=16 and C=12)

I. All CO is used

II. 28g Fe is formed

III. Total mass in the container is 82g

Solution:

Molar mass of;

Fe2O3=2.(56) + 3(16)=160g/mol

CO=12+16=28g/mol

mole of Fe2O3 is;

nFe2O3=40/160=0,25mol

I. If we assume all CO is used;

3 mol CO requires 1mol Fe2O3

1,5mol CO requires ?mol Fe2O3

—————————————————

?=0,5mol Fe2O3, however we have only 0,25mol Fe2O3 so, all CO is not used in this reaction.I is false.

II. All Fe2O3 is used in this reaction.

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

1mol         3mol          2mol      3mol  (for 1 mole Fe2O3)

0,25mol    1,5mol        ?            ?     (We have)

-0,25mol   -0,75mol    +0,50mol +0,75mol (used and formed)

——————————————————————————————————

Thus, 0,75mol CO is used and 0,50mol Fe and 0,75mol CO2 are formed.

Mass of Fe=0,5.(56)=28g II is true.

III. Conservation of mass law states that, amount of matter stays constant before and after reaction. So,

Mass of CO before reaction=42g

Mass of Fe2O3 before reaction=40

Total mass of matters=42 + 40=82g III is true.

4. Find types of given reactşons below.

I. H2(g) + 1/2O2(g) → H2O(l)

II. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

III. CaCO3(s) → CaO(s) + CO2(g)

Solution:

I. It is combustion and formation reaction.

II. It is acid-base reaction or neutralization reaction.

III. It is analysis or decomposition reaction.

5. Which one of the following statements is false for following reactions,

KClO3(s) + Heat → KCl(s) + 3/2O2(g)

N2(g) + 3H2(g) → 2NH3(g) + Heat

C(s) + O2(g) → CO2(g) + Heat

I. First reaction is analysis and others are synthesis reactşons.

II. First one is endothermic and others are exothermic reactions

III. First and third reactions are heterogeneous.

IV. In second reaction volume increases after reaction.

Solution:

I is true first one is analysis reaction and second and third are formation reaction.

II. Since first one takes  heat it is endothermic reaction and others give heat, so they are exothermic reactions.II is true.

III. There are different phases of matters in first and third reactions, they are heterogeneous reactions. III is true.

IV. We write volumes of matters and see whether it increases or decreases;

N2(g) + 3H2(g) → 2NH3(g)

V         3V              2V

Volume of reactants= V + 3V =4V

Volume of products=2V

So, IV is false, volume does not increases.On the contrary, it decreases.

The Original Author: