To react with a #5.00*g# mass of calcium hydroxide, what volume of #0.100*mol*L^-1# #HCl# is required?

Question

Serenity Allen · Accepted Answer

Approx. #1.4*L# hydrochloric acid are required.

We need (i) a chemical reaction: #Ca(OH)_2(s) + 2HCl(aq)rarr CaCl_2(aq) + 2H_2O(l)#* And (ii), the molar quantity of calcium hydroxide: #=# #(5.00*g)/(74.09*g*mol^-1)# #=# #0.0675*mol#. Given the rxn stoichiometry, clearly, we need #2xx0.0675*mol=0.135*mol# #HCl(aq)#. And given our concentration of #HCl#, #(0.135*mol)/(0.100*mol*L^-1)xx10^3*mL*L^-1# #=# #??mL#

Noah Aldrich · Answer

undefined To calculate the volume of 0.100 mol/L HCl required to react with 5.00 g of calcium hydroxide (Ca(OH)2), we first need to determine the moles of Ca(OH)2 present in 5.00 g:

1. Calculate the molar mass of Ca(OH)2:
   Ca: 1 × 40.08 g/mol = 40.08 g/mol
   O: 2 × 16.00 g/mol = 32.00 g/mol
   H: 2 × 1.01 g/mol = 2.02 g/mol
   Total molar mass: 40.08 + 32.00 + 2.02 = 74.10 g/mol

2. Calculate the moles of Ca(OH)2:
   moles = mass / molar mass
   moles = 5.00 g / 74.10 g/mol ≈ 0.0675 mol

Since the balanced chemical equation for the reaction between Ca(OH)2 and HCl is:

Ca(OH)2 + 2HCl -> CaCl2 + 2H2O

we see that 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Therefore, the moles of HCl required will be twice the moles of Ca(OH)2:

moles of HCl = 2 × 0.0675 mol = 0.135 mol

Finally, we can calculate the volume of 0.100 mol/L HCl required:

volume = moles / concentration
volume = 0.135 mol / 0.100 mol/L = 1.35 L

Therefore, 1.35 liters of 0.100 mol/L HCl is required to react with 5.00 g of calcium hydroxide.