Question

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

Answer 1

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

Answer 2

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