Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate.
`NaHCO_(3_((aq)))+HCl_((aq)) to NaCl_((eq))+H_(2)O_((l))+CO_(2_(g))`
How many milliliters of 0.125 M `NaHCO_(3)` solution are needed to neutralize 18.0 mL of 0.100 M HCI?

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.

Stomach acid, a dilute solution of HCl can be neutralised by reaction with aluminig hydroxide Al (OH)_(3) + 3HCl (aq) to AlCl _(3) + 3H _(2)O How many milliliters of 0.1 M Al (OH) _(3) solution are needed to neutralise 21 mL of 0.1 M HCl ?

Stomach acid a dilute solution of HCI can be neutralised by reaction with Aluminium hydroxide. Al(OH)_3 +3HCl(aq) to AlCl_3 +3H_2O how many millitres of 0.1 M Al(OH)_3 solution are needed to neutralise 21 ml of 0.1 M HCl?

What volume of 0.00300 M Hcl solution would just neutralize 30.0 mL of 0.00100 M Ca(OH)_(2) solution?

What mass of NaHCO_(3) (M=84.0) is required to completely neutralize 25.0 mL of 0.125 M H_(2)SO_(4) ?

Find the [OH^-] in 100 mL of 0.015 M HCl (aq.) solution.

What volume of 0.250 M HCl (aq) is required to react completely with 22.6 g of sodium carbonate according to the reaction : Na_(2)CO_(3)(s) + 2HCl (aq) to 2NaCl(aq) + H_(2)O + CO_(2)