How do you balance #MgO + Fe - Fe_2O_3 + Mg#?

Question

How do you balance #MgO + Fe -> Fe_2O_3 + Mg#?

Natalie Anton · Accepted Answer

# 3 MgO + 2 Fe = Fe_2O_3 + 3 Mg#

To form Iron II Oxide, two Fes must be present in the reactants to supply the two Fes, three MgOs must be present in the reactants to supply the three Os, and if three MgOs are present, then three Mgs must be present in the products.

Noah Aldrich · Answer

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

#"Magnesium oxide"# is REDUCED to #"magnesium metal:"# #stackrel(II^+)MgO + 2H^+ +2e^(-) rarr stackrel(0)Mg(s) +H_2O(l)# #(i)# Iron metal is converted to ferric oxide through oxidation. #stackrel(0)"Fe" + 3H_2O(l)rarr stackrel(III^+)"Fe"_2"O"_3(s)+ 6H^+ +6e^(-)# #(ii)#... ...and so we take #3xx(i)+(ii)# to get: #3MgO + 6H^+ +6e^(-) +Fe + 3H_2O(l)rarr 3Mg(s) +"Fe"_2"O"_3(s)+ 6H^+ +6e^(-)+3H_2O(l)# And after the cancellation... #"Fe+ 3MgO" rarr "3Mg(s) +" "Fe"_2"O"_3(s)# Mass and charge are balanced as needed.

Ellie Andrews · Answer

#3MgO+ 2Fe rarrFe_2O_3+Mg#

#color(blue)(MgO+ Fe rarrFe_2O_3+Mg#

To balance the equation, all the elements' respective numbers of molecules must be equal on both sides.

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

#rarr3MgO+ Fe rarrFe_2O_3+Mg#

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

#rarr3MgO+ Fe rarrFe_2O_3+3Mg#

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

#rarr3MgO+ 2Fe rarrFe_2O_3+3Mg#

Now that every element in the equation has an equal number of atoms on both sides, the equation is balanced.

#color(purple)(3MgO+ 2Fe rarrFe_2O_3+Mg#

I hope that's helpful! 😣

Owen Ambrose · Answer

undefined MgO + Fe -> Fe₂O₃ + Mg is already balanced.

How do you balance #MgO + Fe -&gt; Fe_2O_3 + Mg#?

How do you balance #MgO + Fe -> Fe_2O_3 + Mg#?