If 43 mL of a 0.208 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? 2HCl (aq) + Ca(OH)2 (aq) → CaCl2 (aq) + 2H2O (l)
Added by Victoria M.
Step 1
First, we need to find the moles of HCl in the solution. We can do this by multiplying the volume (in liters) by the molarity: moles of HCl = 0.043 L * 0.208 mol/L = 0.008944 mol Show more…
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