How many liters of. 100 m hcl hcl would be required to react with 5 grams of calcium hydroxide?

The number of liters of 0.100 HCl that would be required to react with 5g of Ca (OH) 2 is calculated as follows

find the moles of Ca (OH) 2 used

= mass/molar mas

= 5g / 74.09 g/mol = 0.0675 moles

write the equation involved

that is Ca (OH) 2 + 2 HCl = CaCl2 + 2 H2O

by use of mole ratio between Ca (OH) 2 to HCl which is 1:2 the moles of HCl is therefore = 2 x 0.0675 = 0.135 moles of HCl

volume of HCl = moles of HCL / molarity of HCl

= 0.135 / 0.100 = 1.35 L

The balanced equation for the reaction between HCl and Ca (OH) ₂ is as follows;

Ca (OH) ₂ + 2HCl - - - > CaCl₂ + 2H₂O

stoichiometry of Ca (OH) ₂ to HCl is 1:2

mass of Ca (OH) ₂ reacting - 5 g

therefore number of moles of Ca (OH) ₂ - 5 g / 74 g/mol = 0.068 mol

according to molar ratio

number of HCl moles reacted = twice the number of Ca (OH) ₂ reacted

number of HCl moles required - 0.068 x 2 = 0.136 mol

molarity if HCl solution - 0.100 M

there are 0.100 mol of HCl in 1 L

therefore 0.136 mol in - 0.136 mol / 0.100 mol/L = 1.36 L

volume of 0.100 M HCl required - 1.36 L