How do you balance the equation: $(A) F e S O_{4} + (B) K M n O + (C) H_{2} S O_{4} \to (D) F e_{2} {(S O_{4})}_{3} + (E) K_{2} S O_{4} + (F) M n S O_{4} + (G) H_{2} O$ $(A)FeSO_4+(B)KMnO+(C)H_2SO_4 -> (D)Fe_2(SO_4)_3+(E)K_2SO_4+(F)MnSO_4+(G)H_2O$ ?

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How do you balance the equation: $(A) F e S O_{4} + (B) K M n O + (C) H_{2} S O_{4} \to (D) F e_{2} {(S O_{4})}_{3} + (E) K_{2} S O_{4} + (F) M n S O_{4} + (G) H_{2} O$ $(A)FeSO_4+(B)KMnO+(C)H_2SO_4 -> (D)Fe_2(SO_4)_3+(E)K_2SO_4+(F)MnSO_4+(G)H_2O$ ?

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Answer 1 · 2016-02-29T21:02:28

$10 F e S O_{4} + 2 K M n O + 8 H_{2} S O_{4} \to 5 F e_{2} {(S O_{4})}_{3} + K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O$ $10FeSO_4+2KMnO+8H_2SO_4 -> 5Fe_2(SO_4)_3+K_2SO_4+2MnSO_4+8H_2O$

Explanation:

Let's eliminate unknowns backwards, starting with $G$ $G$ .

From the equations for $H$ $H$ , $M n$ $Mn$ , $K$ $K$ and $F e$ $Fe$ we find:

$G = C$ $G=C$

$F = B$ $F=B$

$E = \frac{1}{2} B$ $E=1/2 B$

$D = \frac{1}{2} A$ $D=1/2 A$

Substituting these in the remaining equations we get:

$S : A + C = 3 (\frac{1}{2} A) + (\frac{1}{2} B) + B = \frac{3}{2} A + \frac{3}{2} B$ $S: A+C=3(1/2 A)+(1/2 B)+B = 3/2A + 3/2B$

$O : 4 A + 4 B + 4 C = 12 (\frac{1}{2} A) + 4 (\frac{1}{2} B) + 4 B + C = 6 A + 6 B + C$ $O: 4A+4B+4C = 12 (1/2 A)+4 (1/2 B)+4B+C = 6A + 6B+C$

Subtracting $A$ $A$ from both sides of the equation for $S$ $S$ we get:

$C = \frac{1}{2} A + \frac{3}{2} B$ $C=1/2A+3/2B$

Substitute this in our equation for $O$ $O$ to get:

$4 A + 4 B + 4 (\frac{1}{2} A + \frac{3}{2} B) = 6 A + 6 B + (\frac{1}{2} A + \frac{3}{2} B)$ $4A+4B+4(1/2A+3/2B) = 6A+6B+(1/2A+3/2B)$

That is:

$6 A + 10 B = \frac{13}{2} A + \frac{15}{2} B$ $6A+10B = 13/2A+15/2B$

Multiplying both sides by $2$ $2$ that becomes:

$12 A + 20 B = 13 A + 15 B$ $12A+20B = 13A+15B$

Subtract $12 A + 15 B$ $12A+15B$ from both sides to get:

$5 B = A$ $5B=A$

So if we put $B = 2$ $B=2$ then we get:

$A = 5 B = 10$ $A=5B=10$

$B = 2$ $B=2$

$C = \frac{1}{2} A + \frac{3}{2} B = 5 + 3 = 8$ $C=1/2A+3/2B=5+3=8$

$D = \frac{1}{2} A = 5$ $D=1/2A = 5$

$E = \frac{1}{2} B = 1$ $E=1/2B = 1$

$F = B = 2$ $F=B=2$

$G = C = 8$ $G=C=8$

So:

$10 F e S O_{4} + 2 K M n O + 8 H_{2} S O_{4} \to 5 F e_{2} {(S O_{4})}_{3} + K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O$ $10FeSO_4+2KMnO+8H_2SO_4 -> 5Fe_2(SO_4)_3+K_2SO_4+2MnSO_4+8H_2O$

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How do you balance the equation: $(A) F e S O_{4} + (B) K M n O + (C) H_{2} S O_{4} \to (D) F e_{2} {(S O_{4})}_{3} + (E) K_{2} S O_{4} + (F) M n S O_{4} + (G) H_{2} O$ $(A)FeSO_4+(B)KMnO+(C)H_2SO_4 -> (D)Fe_2(SO_4)_3+(E)K_2SO_4+(F)MnSO_4+(G)H_2O$ ?

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How do you balance the equation: (A)FeSO4+(B)KMnO+(C)H2SO4→(D)Fe2(SO4)3+(E)K2SO4+(F)MnSO4+(G)H2O(A)FeSO_4+(B)KMnO+(C)H_2SO_4 -> (D)Fe_2(SO_4)_3+(E)K_2SO_4+(F)MnSO_4+(G)H_2O ?

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How do you balance the equation: $(A) F e S O_{4} + (B) K M n O + (C) H_{2} S O_{4} \to (D) F e_{2} {(S O_{4})}_{3} + (E) K_{2} S O_{4} + (F) M n S O_{4} + (G) H_{2} O$ $(A)FeSO_4+(B)KMnO+(C)H_2SO_4 -> (D)Fe_2(SO_4)_3+(E)K_2SO_4+(F)MnSO_4+(G)H_2O$ ?