Neutrakization of acid by base: Stomach acid, a dilute solution of HCl in water can be neutralized by reaction with sodium hydrogen carbonate, `NaHCO_(3)`, according to the equation
`HCl(aq.)+NaHCO_(3)(aq.) rarr NaCl(aq.)+H_(2)O(l)+CO_(2)(g)`
How many millilitres of 0.125 M `NaHCO_(3)` solution are needed to neutralize 18.0 mL of 0.100 M HCl?
Strategy: Solving stoichiometry problems always requires finding the number of moles of the first reactant. using the coefficients of the balanced chemical equation to find the number of moles of the second reactant, and then finding the amount of the second reactant. The flow diagram in Figure 2.1 summarizes the situation.

Explanation

Step 1. Find moles of HCl present in `18.0 mL` of a `0.100 M `solution by multiplying volume with molarity:
Moles of `HCl=V_(L)xxM`
`=(18.0 mL)((1 L)/(1000 mL))(0.100(mol)/L)`
`=0.0018 mol HCl`
`=1.80xx10^(-3) mol HCl`
Step 2. According to the coefficients of the balanced equation each mole of HCl reacts with 1 mol of `NaHCO_(3)`. Thus `1.80xx10^(-3)` mole HCl will be neutralized by `1.80xx10^(-3)` mol `NaHCO_(3)`.
Step 3. Find milititres of `0.125 M NaHCO_(3)` solution containing `1.80xx10^(-3)` mol `NaHCO_(3)` by dividing moles by molarity:
Volume of `NaHCO_(3) soln= ("Moles of" NaHCO_(3))/("Molarity of" NaHCO_(3) soln)`
`=(1.80xx10^(-3) mol NaHCO_(3))/(0.125 mol NaHCO_(3)//L soln)`
=0.0144 L soln
=14.4 mL soln
Thus, `14.4 mL` of the `0.125 M NaHCO_(3)` solution is needed to neutralize `18.0 mL` of the `0.100 M` HCl solution.