Asked Dec 2, 2019

Find the volume of 0.110 M hydrochloric acid necessary to react completely with 1.74 g Al(OH)3


Expert Answer

Step 1


Mass of Al(OH)3 = 1.74 g

Molarity of HCl = 0.110 M = 0.11 mol/L

Moles of Al(OH)3  can be calculated as:


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Molar mass of Al(OH)3 27 3(16 1) 78 g/mol 1.74 g Mass Moles 0.0223 mol Molar mass 78 g/mol

Step 2

The balanced reaction of HCl and Al(OH)3 is:

3 HCl + Al(OH)3 → AlCl3 + 3 H2O

From the reaction it is evident that:


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1 mol Al(OH)3 requires 3 mol HCl 3 mol HCl . 0.0233 mol Al(OH)s requires x 0.0233 mol Al(OH)3 1 mol Al (OH)3 = 0.0699 mol HCI

Step 3

Volume of HCl required can...


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Moles MolarityVolume Moles Volume Molarity 0.0699 mol Volume 0.11 mol/L Volume 0.6355 L Volume 635.5 mL (1 L 1000 mL)


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