For the redox reaction, 
MnO4-+C2O42-+H+Mn2++CO2+H2O, the correct coefficients of the reactants  MnO4-,C2O4-,H+ for the balanced reaction are, respectively:

1. 2, 5, 16

2. 16, 3, 12

3. 15, 16, 12

4. 2, 16, 5

Subtopic:  Balancing of Equations |
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The coefficient of I-\(IO_{3}^{-}\) and H+ in the redox reaction;  I- + IO3- + H+ → I2 + H2O  in the balanced form respectively are:

1. 5, 1, 6

2. 1, 5, 6

3. 6, 1, 5

4. 5, 6, 1

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When KMnO4 acts as an oxidizing agent and ultimately forms \(\left[\mathrm{MnO}_4\right]^{-2}, \mathrm{MnO}_2, \mathrm{Mn}_2 \mathrm{O}_3, \mathrm{Mn}^{+2}\)
then the number of electrons transferred in each case, respectively, are:

1. 4, 3, 1, 5 2. 1, 5, 3, 7
3. 1, 3, 4, 5 4. 3, 5, 7, 1
Subtopic:  Balancing of Equations |
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In a balanced equation
H2SO4+xHIH2S+yI2+zH2O, the values of x, y, z are:

1. x = 3, y = 5, z = 2
2. x = 4, y = 8, z = 5
3. x = 8, y = 4, z = 4
4. x = 5, y = 3, z = 4
Subtopic:  Balancing of Equations |
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The reaction, 
5H2O2+XClO2+2OHXCl+YO2+6H2O 
is balanced if:
1. 𝑋 = 5, 𝑌 = 2
2. 𝑋 = 2, 𝑌 = 5
3. 𝑋 = 4, 𝑌 = 10
4. 𝑋 = 5, 𝑌 = 5

Subtopic:  Balancing of Equations |
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Consider the following equation:
\(\small[Cr_2O^{2-}_7 (aq) + 3SO^{2-}_3 (aq) + 8H^+(aq) \rightarrow A + B + 4H_2O(l)\)

The product A and B are, respectively:

1. \(2 \mathrm{Cr}^{2+} ; 3 \mathrm{SO}_2 \)
2. \(2 \mathrm{Cr}^{+} ; \mathrm{S}_2 \mathrm{O}_7{ }^{2-} \)
3. \(2 \mathrm{Cr}^{3+} ; 3 \mathrm{SO}_4{ }^{2-} \)
4. \(2 \mathrm{Cr}^{3+} ; 3 \mathrm{HSO}_4{ }^{-}\)

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For a given reaction scheme;
 \(x \mathrm{Cl}_{2}+y~\mathrm{NaOH} \longrightarrow \mathrm{zNaCl}+a~\mathrm{} \mathrm{NaClO}_{3} + bH_2O\)

The ratio of \({ x + y \over z + a+b}\)will be: 
1. One (1)  2. Two (2) 
3. Five (5) 4. Three (3)
Subtopic:  Balancing of Equations |
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In the following reaction, \(X, Y\) and \(Z\) are respectively:
\(XZn+ NO^- _3+ YH^+ \rightarrow XZn^{2+}+ NH^+ _4+ ZH_2O \)

1. 4, 10, and 3
2. 3, 8, and 3
3. 3, 10, and 3
4. 4, 3, and 10
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What is the minimum number of electrons needed to balance the following half-reaction with whole-number coefficients?
\(\text{IO}^-_3\rightarrow \text{I}_2\)

1. Six (6)
2. Four (4)
3. Ten (10)
4. Eight (8)
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When the equation for the reaction given below is balanced with the lowest whole-number coefficients, then the coefficient for \(\mathrm {H}_2\mathrm{O}\) is:
\(\mathrm {FeCl}_2 + \mathrm {KMnO_4} + \mathrm{HCl}~~ \rightarrow \mathrm {FeCl_3} + \ \mathrm{MnCl}_2 + ~\mathrm {H}_2\mathrm {O}\)

1. One
2. Two
3. Three
4. Four
Subtopic:  Balancing of Equations |
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