Respuesta :
Answer
The moles of water that will be produced = 0.343 mol
Explanation
Given:
Mass of Ca(OH)2 = 29.0 g
Mass of HCl = 12.5 g
We know the reaction : Ca(OH)2 + 2HCI —> CaCI2 + 2H2O
Molar mass of Ca(OH)2 = 74.093 g/mol
Molar mass of HCl = 36.458 g/mol
Molar mass of H2O = 18.015 g/mol
Required: Moles of water that will be formed
Solution:
Use the stoichiometry to find the moles of water using both the reactants.
For Ca(OH)2:
[tex]\begin{gathered} 29.0\text{ g Ca\lparen OH\rparen}_2\text{ x }\frac{1\text{ mole Ca\lparen OH\rparen}_2}{74.093\text{ g Ca\lparen OH\rparen}_2}\text{ x }\frac{2\text{ mol H}_2O}{1\text{ mol Ca\lparen OH\rparen}_2} \\ \\ =\text{ 0.783 mol H}_2O \end{gathered}[/tex]For HCl
[tex]\begin{gathered} 12.5\text{ g HCl x }\frac{1\text{ mol HCl}}{36.458\text{ g}}\text{ x }\frac{2\text{ mol H}_2O}{2\text{ mol HCl}} \\ \\ =\text{ 0.343 mol H}_2O \end{gathered}[/tex]HCl will produce less moles of H2O, thus HCl is the limiting reactant/reagent and the moles of water that will be produced = 0.343 mol
