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Chemistry
Chemistry
Problem
Find the smallest whole-number coefficient that must be placed in front of
F
e
2
+
\mathrm{Fe^{2+}}
F
e
2
+
so that charge is conserved in the equation below. Do not change any other coefficients.
M
n
O
4
−
+
8
H
+
+
‾
F
e
2
+
→
M
n
2
+
+
‾
F
e
3
+
+
4
H
2
O
\mathrm{MnO_4^- + 8\,H^+ + \underline{\quad}\,Fe^{2+} \;\rightarrow\; Mn^{2+} + \underline{\quad}\,Fe^{3+} + 4\,H_2O}
Mn
O
4
−
+
8
H
+
+
F
e
2
+
→
M
n
2
+
+
F
e
3
+
+
4
H
2
O
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