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# Help solving this chemistry question watch

1. 6 A white powder is known to be a mixture of magnesium oxide and aluminium oxide.
100 cm3 of 2 mol dm–3 NaOH(aq) is just sufficient to cause the aluminium oxide in x grams of the
mixture to dissolve.
The reaction occurring is Al 2O3 + 2OH– + 3H2O → 2Al (OH)4-
800 cm3 of 2 mol dm–3 HCl (aq) is just sufficient to cause all of the oxide in x grams of the mixture
to dissolve.
The reactions occurring are Al 2O3 + 6H+ → 2Al3+ + 3H2O
and MgO + 2H+ → Mg2+ + H2O.
How many moles of each oxide are present in x grams of the mixture?

aluminium magnesium
oxide oxide
A 0.05 0.25
B 0.05 0.50
C 0.10 0.25
D 0.10 0.50
How do you do this? o_O
Ans is D

An elaborate explanation would be much appreciated
2. Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.

Just quoting in Amusing Elk so she can move the thread if needed
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(Original post by Amusing Elk)
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3. Moles of NaOH 100/1000 x 2= 0.2As. 2 moles NaOH = 1mol Al2O3So. 0.2 mil = 0.1 mil Al2O3Now moles of HCl = 800/1000 * 2 = 1.6As. 1mol Al2O3 = 6 mol HClSo 0.1 mol. =. 0.6 mol HClMol of HCl left. = 1 molAs 2 mol HCl = 1 mol MgO 1 mol HCl = 0.5 mol MgO
4. [QUOTE=Naseerpr;77869904]Moles of NaOH 100/1000 x 2= 0.2As. 2 moles NaOH = 1mol Al2O3So. 0.2 mol = 0.1 mol Al2O3Now moles of HCl = 800/1000 * 2 = 1.6As. 1mol Al2O3 = 6 mol HClSo 0.1 mol. =. 0.6 mol HClMol of HCl left. = 1 molAs 2 mol HCl = 1 mol MgO 1 mol HCl = 0.5 mol MgO

Hope now it will be clear for you...

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