Question

11 balance the following two equations: remember both sides of the equation must have an equal number of atoms. you cannot change the subscripts, only the coefficients.
na: 1 na: 2
h: 1 h: 2
c: 1 c: 2
o: 3 o: 6
a. nahco₃→na₂co₃ + h₂o + co₂
k: 1 k: 1
cl: 1 cl: 1
o: 3 o: 2
b. kclo₃→kcl + o₂

Explanation:

Step1: Balance equation a

For \(NaHCO_3
ightarrow Na_2CO_3 + H_2O+CO_2\), we can see that on the left - hand side (LHS) there is 1 \(Na\) atom and on the right - hand side (RHS) there are 2 \(Na\) atoms. So we put a coefficient of 2 in front of \(NaHCO_3\).
\(2NaHCO_3
ightarrow Na_2CO_3 + H_2O+CO_2\)
Now, count the number of atoms:

• \(Na\): \(2\times1 = 2\) (LHS) and \(2\) (RHS)
• \(H\): \(2\times1=2\) (LHS) and \(2\) (RHS) (\(H_2O\) has 2 \(H\) atoms)
• \(C\): \(2\times1 = 2\) (LHS) (\(NaHCO_3\) has 1 \(C\) atom) and \(1 + 1=2\) (RHS) (\(Na_2CO_3\) has 1 \(C\) atom and \(CO_2\) has 1 \(C\) atom)
• \(O\): \(2\times3=6\) (LHS) (\(NaHCO_3\) has 3 \(O\) atoms) and \(3 + 1+2 = 6\) (RHS) (\(Na_2CO_3\) has 3 \(O\) atoms, \(H_2O\) has 1 \(O\) atom and \(CO_2\) has 2 \(O\) atoms)

Step2: Balance equation b

For \(KClO_3
ightarrow KCl+O_2\), we use the least - common - multiple method for oxygen. The number of \(O\) atoms on the LHS is 3 and on the RHS is 2. The least common multiple of 2 and 3 is 6.
Multiply \(KClO_3\) by 2 and \(O_2\) by 3: \(2KClO_3
ightarrow KCl + 3O_2\)
Now, for \(K\) and \(Cl\), we put a coefficient of 2 in front of \(KCl\)
\(2KClO_3
ightarrow 2KCl+3O_2\)
Check the atoms:

• \(K\): \(2\) (LHS) and \(2\) (RHS)
• \(Cl\): \(2\) (LHS) and \(2\) (RHS)
• \(O\): \(2\times3 = 6\) (LHS) and \(3\times2=6\) (RHS)

Answer:

a. \(2NaHCO_3
ightarrow Na_2CO_3 + H_2O+CO_2\)
b. \(2KClO_3
ightarrow 2KCl+3O_2\)