Answer
a. Na + O2 -> Na2O2
b. P4 + 5O2 -> P4O10
c. Ca(HCO3)2 + Na2CO3 -> CaCO3 + 2NaHCO3
d. 4NH3 + 5O2 -> 4NO + 6H2O
e. 2Rb + 2H2O -> 2RbOH + H2
Work Step by Step
a. Na + O2 -> Na2O2
Both sides already have 2 O atoms.
Balance Na by adding coefficient “2” to the reactant side such that 2x1=2, which is the number of Na atoms in the product side.
2Na + O2 -> Na2O2
b. P4 + O2 -> P4O10
Both sides already have 4 P atoms. Balance O by adding coefficient “5” to O2 so that 5x2=10 O atoms, which is the number of O atoms in product side.
P4 + 5O2 -> P4O10
c. Ca(HCO3)2 + Na2CO3 -> CaCO3 + NaHCO3
Both Ca and CO3 are atoms are balanced. Balance Na and HCO3 by adding coefficient 2 to NaHCO3 such that 2x1=2, which is the number of Na and HCO3 atoms in reactant side.
Ca(HCO3)2 + Na2CO3 -> CaCO3 + 2NaHCO3
d. NH3 + O2 -> NO + H2O
Balance H by looking for the LCM of the given number of atoms. The LCM of 3 and 2 is 6. To get 6 H atoms for both sides, add coefficient “2” to NH3 and “3” to H2O.
2NH3 + O2 -> NO + 3H2O
H is balanced. To balance N, add the coefficient “2” to NO.
2NH3 + O2 -> 2NO + 3H2O
N is balanced. To balance O, multiply O2 by 5/2 such that 5/2x2=5, which is the number of O atoms in product side.
2NH3 + 5/2O2 -> 2NO + 3H2O
O is balanced. To get whole number coefficients, multiply the whole equation by 2.
4NH3 + 5O2 -> 4NO + 6H2O
e. 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
