[AWK]

This would reduce the volume, and you have more volume than would be assumed that the gas is ideal. This allows you to exclude one more option

[kahynickel]

A white powder is known to be a mixture of MgO and Al2O3.

100 cm3 of 2 M NaOH(aq) is just sufficient to cause the Al2O3 in x grams of the mixture to dissolve.

the reaction occurring is
Al2O3 +   2OH- + 3H2O ..2Al(OH)4-

800 cm3 of 2 M HCl(aq) is just sufficient to cause all the oxide in x grams of the mixture to dissolve.

AL2O3   +    6H+ ...2Al3+   +    3H2O
MgO      +    2H+ ...Mg2+    +  H2O

how many moles of each oxide are present in x grams of the mixture?

Answer is :   Al2O3  ..0.10 g
                   MgO ..0.50 g


For moles of Al2O3....According to first equation moles of NaOH = 0.20 ; so moles of AL2O3 will be 0.10

for moles of MgO...There are 1.6 moles of HCl but this is used for BOTH oxides. so will it be

Moles of MgO = [Total moles of HCl] – [Moles of HCl used for Al2O3] /2 . but did not get the right answer ??

[AWK]

Show your result. The given answer in grams is wrong.

[kahynickel]

it's 0.10 moles (Al2O3) and 0.50 moles (MgO).

[kahynickel]

1.6 moles of HCl are available..

Al2O3  :  H+
1         :  6

if 0.1 moles of Al2O3 then 0.6 moles of HCl are used up....leaving 1.00 moles to be reacted with MgO

MgO   :   H+
1        :   2
 x       :1.00 ..x= 0.50.....Is this correct working ??

[AWK]

Answer in moles is correct