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nserant · May 29, 2009, 08:40 PM

Find the mass of AlCl3 that is produced when 510 grams of Al2O3 react with HCl according to the following equation: Al2O3+6HCl=2AlCl3+3H2O .

nserant · May 29, 2009, 08:40 PM

Find the mass of AlCl3 that is produced when 510 grams of Al2O3 react with HCl according to the following equation: Al2O3+6HCl=2AlCl3+3H2O .

Perito · May 30, 2009, 05:33 AM

Originally Posted by nserant

Find the mass of AlCl3 that is produced when 510 grams of Al2O3 react with HCl according to the following equation: Al2O3+6HCl=2AlCl3+3H2O .

You are already given a balanced chemical equation. That would be the first step if you didn't know it:

$Al_2O_3+6HCl=2AlCl_3+3H_2O$

You have 510 grams of $Al_2O_3$ . In the chemical world, we deal in moles when doing this type of calculation. To convert from moles to grams or vice-versa, we need the molecular weight of $Al_2O_3$

Molecular Weight Al2O3 = 2 * atomic weight of Al + 3 * Atomic Weight of O

Molecular Weight Al2O3 ≈ 2 * 27 + 3 * 16 = 102 g/mole.

Given 510 grams, how many moles of Al2O3 do we have?

$510\,g \times \frac {1\,mole}{102\,g} =5\,moles$

From the balanced chemical equation, we can see that 2 moles of AlCl3 are produced for every mole of Al2O3 that reacts. Therefore, we'll produce 2 x 5 = 10 moles of AlCl3.

Molecular Weight AlCl3 = 1 * atomic weight of Al + 3 * atomic weight of Cl

Molecular Weight AlCl3 = 27 + 3 * 35.5 = 133.5 g/mole

Since 10 moles of AlCl3 are produced, we can calculate the weight

$10\,moles\,AlCl_3 \times \frac {133.5\,g}{mole} = 1335\,g\,AlCl_3$

Unknown008 · May 30, 2009, 05:44 AM

Was going to answer... great post Perito, had to spread the rep though... :( Nice avy! :D Jerry's got a new friend now :D:D:D