Question

A white powder is known to be a mixture of magnesium oxide and aluminum oxide. 100cm3 of 2moldm-3 NaOH(aq) is just sufficient to cause the aluminum oxide in x grams of the mixture to dissolve. The reaction occurring is Al2O3 + 2OH- + 3H2O ----> 2Al(OH)4- 800cm3 of 2 moldm-3 HCl (aq) is just sufficient to cause all of the oxide in x grams of the mixture to dissolve. The reactions occurring are Al2O3 + 6H ---> 2Al3+ + 3H2O and Mg2+ +H2O. How many moles of each oxide are present in x grams of the mixture ?
A. Aluminium Oxide 0.05, Magnesium Oxide 0.25

B. Aluminium Oxide 0.05, Magnesium Oxide 0.50

C.Aluminium Oxide 0.10, Magnesium Oxide 0.25

D.Aluminium Oxide 0.10, Magnesium Oxide 0.50

Answer 1

Answer: D.Aluminium Oxide 0.10, Magnesium Oxide 0.50

Explanation:

Number of moles of NaOH= number of moles × volume

Number of moles= 100/1000 × 2 = 0.2 moles

Since;

2 moles of NaOH yield 1 mole of Al2O3

0.2 moles of NaOH will yield 0.2 × 1/2 = 0.1 moles of Al2O3.

Number of moles of HCl= 800/1000 × 2 = 1.6 moles

If 1 mole of Al2O3 requires 6 moles of HCl

0.1 moles of Al2O3 requires 0.1 × 6 = 0.6 moles of HCl.

Number of moles of HCl left after reaction with Al2O3 = 1.6- 0.6 = 1 mole

This leftover reacts with MgO

But;

1 mole of MgO reacts with 2 moles of HCl

x moles of MgO reacts with 1 mole of HCl

Thus; x= 0.5 moles of MgO