If 43.0 mL of 0.220 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution? Express the mass in grams to three significant digits.
Guide On Rating System
Vote
The balanced chemical equation for the reaction between HCl and Ca(OH)2 is:
2HCl + Ca(OH)2 → CaCl2 + 2H2O
From the equation, we can see that 2 moles of HCl react with 1 mole of Ca(OH)2. Therefore, the molar ratio between HCl and Ca(OH)2 is 2:1.
To find the moles of HCl, we can use the formula:
moles = volume (in L) × molarity
Given that the volume of HCl is 43.0 mL (or 0.0430 L) and the molarity is 0.220 M, we can calculate the moles of HCl:
moles HCl = 0.0430 L × 0.220 M = 0.00946 moles HCl
Using the molar ratio, we know that there will be half that amount of moles of Ca(OH)2 since the ratio is 2:1. Therefore, the moles of Ca(OH)2 is:
moles Ca(OH)2 = 0.00946 moles HCl × 1/2 = 0.00473 moles Ca(OH)2
To find the mass of Ca(OH)2, we can use its molar mass, which is calculated by adding up the atomic masses of calcium (Ca), oxygen (O), and hydrogen (H):
molar mass Ca(OH)2 = (40.08 g/mol + 2 × 16.00 g/mol + 2 × 1.01 g/mol) = 74.09 g/mol
Now we can calculate the mass of Ca(OH)2:
mass Ca(OH)2 = moles Ca(OH)2 × molar mass Ca(OH)2
mass Ca(OH)2 = 0.00473 moles × 74.09 g/mol = 0.352 g
Therefore, there must be approximately 0.352 grams of Ca(OH)2 in the solution.
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99