How do you balance $M g O + F e \to F e_{2} O_{3} + M g$ $MgO + Fe -> Fe_2O_3 + Mg$ ?

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How do you balance $M g O + F e \to F e_{2} O_{3} + M g$ $MgO + Fe -> Fe_2O_3 + Mg$ ?

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Answer 1 · 2018-05-24T02:11:29

$3 M g O + 2 F e = F e_{2} O_{3} + 3 M g$ $3 MgO + 2 Fe = Fe_2O_3 + 3 Mg$

Explanation:

There must be 2 Fe's in the reactants to supply the 2 Fe's needed to form the Iron II Oxide.
There must be 3 MgO 's in the reactants to supply the 3 O's needed to form the Iron II Oxide.
If there are 3 MgO's then there are three Mg's in the products.

Answer 2 · 2018-05-24T08:07:34

$3 M g O + 2 F e \to F e_{2} O_{3} + M g$ $3MgO+ 2Fe rarrFe_2O_3+Mg$

Explanation:

$M g O + F e \to F e_{2} O_{3} + M g$ $color(blue)(MgO+ Fe rarrFe_2O_3+Mg$

Balancing the equation means, making the number of molecules of all the different elements in the equation equal on both sides.

First, lets check $O$ $O$ (Oxygen). It has $1$ $1$ atom in the left and $3$ $3$ atoms in the right. So, multiply the left hand $M g O$ $MgO$ by $3$ $3$

$\to 3 M g O + F e \to F e_{2} O_{3} + M g$ $rarr3MgO+ Fe rarrFe_2O_3+Mg$

Now, $M g$ $Mg$ has $3$ $3$ atoms in the left and $1$ $1$ atom on the right.
So, multiply the right side $M g$ $Mg$ by $3$ $3$

$\to 3 M g O + F e \to F e_{2} O_{3} + 3 M g$ $rarr3MgO+ Fe rarrFe_2O_3+3Mg$

Now, $F e$ $Fe$ has $1$ $1$ atom on the left and $2$ $2$ atoms in the right, So, multiply the $F e$ $Fe$ in the left hand side by $2$ $2$

$\to 3 M g O + 2 F e \to F e_{2} O_{3} + 3 M g$ $rarr3MgO+ 2Fe rarrFe_2O_3+3Mg$

Now, all the elements in the equation have equal atoms in both sides. So it is now a balanced equation

$3 M g O + 2 F e \to F e_{2} O_{3} + M g$ $color(purple)(3MgO+ 2Fe rarrFe_2O_3+Mg$

Hope that helps! ☺

Answer 3 · 2018-05-26T13:43:42

We could split this equation into individual redox reactions....

Explanation:

$Magnesium oxide$ $"Magnesium oxide"$ is REDUCED to $magnesium metal:$ $"magnesium metal:"$

$I I^{+} M g O + 2 H^{+} + 2 e^{-} \to 0 M g (s) + H_{2} O (l)$ $stackrel(II^+)MgO + 2H^+ +2e^(-) rarr stackrel(0)Mg(s) +H_2O(l)$ $(i)$ $(i)$

$Iron metal$ $"Iron metal"$ is OXIDIZED to $ferric oxide...:$ $"ferric oxide...:"$

$0 Fe + 3 H_{2} O (l) \to {I I I^{+} Fe}_{2} O_{3} (s) + 6 H^{+} + 6 e^{-}$ $stackrel(0)"Fe" + 3H_2O(l)rarr stackrel(III^+)"Fe"_2"O"_3(s)+ 6H^+ +6e^(-)$ $(i i)$ $(ii)$ ...

...and so we take $3 \times (i) + (i i)$ $3xx(i)+(ii)$ to get:

$3 M g O + 6 H^{+} + 6 e^{-} + F e + 3 H_{2} O (l) \to 3 M g (s) + {Fe}_{2} O_{3} (s) + 6 H^{+} + 6 e^{-} + 3 H_{2} O (l)$ $3MgO + 6H^+ +6e^(-) +Fe + 3H_2O(l)rarr 3Mg(s) +"Fe"_2"O"_3(s)+ 6H^+ +6e^(-)+3H_2O(l)$

….and upon cancellation...

$Fe+ 3MgO \to 3Mg(s) + {Fe}_{2} O_{3} (s)$ $"Fe+ 3MgO" rarr "3Mg(s) +" "Fe"_2"O"_3(s)$

Charge and mass are balanced as required....

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How do you balance $M g O + F e \to F e_{2} O_{3} + M g$ $MgO + Fe -> Fe_2O_3 + Mg$ ?

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How do you balance $M g O + F e \to F e_{2} O_{3} + M g$ $MgO + Fe -> Fe_2O_3 + Mg$ ?