Consider the following reaction: 2hcl + ca(oh)2 cacl2 + 2h2o a scientist wants to neutralize 25 ml of 0.001 m ca(oh)2 with a volume of 0.005 m hcl. what is the minimum volume of hcl required?

The balanced chemical equation for the above reaction is as follows ;
Ca(OH)2 + 2HCl —> CaCl2 + 2H2O
Stoichiometry of Ca(OH)2 to HCl is 1:2
Number of Ca(OH)2 moles present - 0.001 mol/L / 1000 mL/L x 25 mL
Number of Ca(OH)2 moles = 2.5 x 10^(-5) mol
Number of HCl moles needed for neutralisation = 2.5 x 10^(-5) mol x2 = 5 x 10 ^(-5) mol
The molarity of HCl solution is 0.005 M
The solution contains 0.005 mol in 1000 mL
Therefore volume of 5x10^(-5) mol in = 1000/0.005 x 5 x 10^(-5) mol = 10 mL
10 mL of HCl is needed for neutralisation

Only one reactant yields 2 or more products. meaning you will only have 1 product and it will break down into 2 or more products. eg. h2o -> h2 + o2