Skip to content
Choose the Best Answer · Q3

Q.Stomach acid, a dilute solution of HCl can be neutralised by reaction with Aluminium hydroxide Al(OH)3+3HCl(aq)→AlCl3+3H2O\mathrm{Al(OH)_3 + 3HCl(aq) \rightarrow AlCl_3 + 3H_2O}. How many millilitres of 0.1 M Al(OH)3_3 solution are needed to neutralise 21 mL of 0.1 M HCl?

(a) 14 mL
(b) 7 mL
(c) 21 mL
(d) none of these
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
4% · 3/67 Questions
✓ Free question

Step 1. The balanced equation is Al(OH)3+3HCl(aq)→AlCl3+3H2O\mathrm{Al(OH)_3 + 3HCl(aq) \rightarrow AlCl_3 + 3H_2O}, so 1 mole of Al(OH)3_3 reacts with 3 moles of HCl.

Step 2. Moles of HCl to neutralise =M×V=0.1 mol/L×0.021 L=2.1×10−3=M\times V = 0.1\ \text{mol/L}\times0.021\ \text{L}=2.1\times10^{-3} mol.

Step 3. Moles of Al(OH)3_3 needed =2.1×10−33=0.7×10−3=\dfrac{2.1\times10^{-3}}{3}=0.7\times10^{-3} mol.

Step 4. Volume of 0.1 M Al(OH)3_3 needed =0.7×10−3 mol0.1 mol/L=7×10−3=\dfrac{0.7\times10^{-3}\ \text{mol}}{0.1\ \text{mol/L}}=7\times10^{-3} L =7=7 mL.

✓Final answer

(b) 7 mL

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.