Answer
a. Fe2O3 + 3CO -> 2Fe + 3CO2
b. 2Rb + 2H2O -> 2RbOH + H2
c. 10K + 2KNO3 -> 6K2O + N2
d. (NH4)2Cr2O7 -> N2 + 4H2O + Cr2O3
e. 2Al + Cr2O3 -> Al2O3 + 2Cr
Work Step by Step
a) Fe2O3 + CO -> Fe + CO2
This equation could only be balanced if CO and CO2 have the same coefficient. The smallest possible coefficient is 3. Balance C and O by adding coefficient “3” to both CO and CO2.
Fe2O3 + 3CO -> Fe + 3CO2
Now you have 6 O and 3 C atoms to both sides. Balance Fe by adding the coefficient “2” to the product side.
Fe2O3 + 3CO -> 2Fe + 3CO2
(b) Rb + H2O -> RbOH + H2
Balance H and O by adding the coefficient “2” to both RbOH and H20.
Rb + 2H2O -> 2RbOH + H2
H & O are balanced. Balance Rb by adding the coefficient “2” to Rb.
2Rb + 2H2O -> 2RbOH + H2
(c) K + KNO3 -> K2O + N2
Balance N by adding the coefficient “2” to KNO3.
K + 2KNO3 -> K2O + N2
Balance O by adding the coefficient “6” in K2O.
K + 2KNO3 -> 6K2O + N2
Finally, balance K atoms by adding the coefficient “10” to K such that 10+2=12, which is the number of K atoms in the product side.
10K + 2KNO3 -> 6K2O + N2
(d) (NH4)2Cr2O7 -> N2 + H2O + Cr2O3
Both N and Cr are already balanced.
Balance H by adding the coefficient “4” to H20 such that 4x2=8, which is the number of H atoms in the reactant side.
(NH4)2Cr2O7 -> N2 + 4H2O + Cr2O3
H and O are balanced.
(e)Al + Cr2O3 -> Al2O3 + Cr
O is already balanced.
Balance Al atoms by adding the coefficient “2” to Al.
2Al + Cr2O3 -> Al2O3 + Cr
Balance Cr atoms by adding the coefficient “2” to Cr.
2Al + Cr2O3 -> Al2O3 + 2Cr
