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

3 Answers
May 24, 2018

Answer:

# 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.

May 24, 2018

Answer:

#3MgO+ 2Fe rarrFe_2O_3+Mg#

Explanation:

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

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

Hope that helps! ☺

May 26, 2018

Answer:

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

Explanation:

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

#stackrel(II^+)MgO + 2H^+ +2e^(-) rarr stackrel(0)Mg(s) +H_2O(l)# #(i)#

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

#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 upon cancellation...

#"Fe+ 3MgO" rarr "3Mg(s) +" "Fe"_2"O"_3(s)#

Charge and mass are balanced as required....