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To determine how many moles of aluminum (Al) are produced from 20 moles of aluminum oxide (Al2O3), we need to use the balanced chemical equation for the reduction of Al2O3. The equation is: 2 Al2O3 → 4 Al + 3 O2. From this, we see that 2 moles of Al2O3 produce 4 moles of Al. Therefore, from 20 moles of Al2O3, we can calculate that 20 moles of Al2O3 would produce 40 moles of Al.

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1d ago

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How many moles of aluminum oxide (Al2O3) are formed in the reaction Fe2O3 plus 2Al and rarr Al2O3 plus 2Fe?

Fe2O3 + 2Al ===> Al2O3 + 2FeIn this reaction the number of moles of Al2O3 produced is dependent on the number of moles of Fe2O3 and Al that one starts with. For every 1 mole Fe2O3 and 2 moles Al, one gets 1 moles of Al2O3.


How many moles of Al203 are formed when 0.78mol o2 reacts with aluminium?

The balanced reaction equation is 4Al + 3O2 -> 2Al2O3. Therefore, 3 moles of O2 reacts with 4 moles of Al to form 2 moles of Al2O3. Since 0.78 mol of O2 is reacted, the number of moles of Al2O3 formed can be calculated using the stoichiometry of the reaction.


How many moles of aluminum oxide ( Al2O3) are formed in the reaction Fe2O3 2Al-Al2O3 2Fe?

In the given reaction, the stoichiometry is 1:1 for Fe2O3 to Al2O3. So the number of moles of Al2O3 formed will be the same as the number of moles of Fe2O3 originally present.


How many grams of Al2O3 are formed from 3 moles of O2?

If the reaction is just between O2 and Al, the balanced equation would be:3O2 + 4Al -> 2Al2O3By using the coefficients of the equation, we see that from 3 moles of O2 we get 2 moles of Al2O3. To find out how much this is in grams, we need to find the molar mass of Al2O3 . This is just the sum of the atomic masses of each atom that makes it up. (Use a periodic table to find atomic masses) So we get:26.98*2+16*3= 102.0One mole equals 102g, so the 2 moles produced by the reaction would amount to 204g.


How many moles og Al are in 2.16 mol of Al2O3?

so you find the ratio of Aluminum to Oxygenin this case it is 2:3and then because the total mole is 2.16for aluminum:2.16/5*2 = 0.864for oxygen:2.16/5*3 = 1.296hope this helped :D


How many moles of Al2O3 are present in 49.3 g of this compound?

To find the number of moles, we need to use the molar mass of Al2O3, which is 101.96 g/mol. Divide the given mass (49.3 g) by the molar mass to get the number of moles of Al2O3 present. 49.3 g / 101.96 g/mol ≈ 0.483 moles of Al2O3.


How many grams of oxygen are in 5.75 moles of aluminum oxide?

Well to find how many grams are in moles you would eventually multiply the mole by the molar mass. The molar mass of aluminum oxide would be 101.96 ( you would find that by multiplying the atomic mass of al by 2 and o by 3 and adding them together). But the molar mass of Oxygen is just about 48 (rounded to 16 instead of 15.9994)5.75 moles of Al2O3 X 48 g oxygen/1 mole of Al2O3=276 g oxygen in 5.75 mole Al2O3


How many moles of aluminum oxide can be made if 5.23 moles Al completely react?

This is a mole stoichiometry problem. Start with the balanced equation for the synthesis of aluminum oxide: 4Al + 3O2 --> 2Al2O3. The ratio of aluminum to aluminum oxide in this equation is 4:2, or 2:1, so 5.23 moles Al means half that number for Al2O3, so about 2.62 moles of aluminum oxide will be produced.


How many moles are in 6.80 g of Al2O3?

Remember the Moles Eq'n. moles = mass(g) / Mr The Mr ( Relative molecular mass of Al2O3) is . 2 x Al = 2 x 27 = 54 3 x O = 3 x 16 = 48 54 + 48 = 102 substituting moles(Al2O3) = 6.8 g / 102 moles = 0.0666.... ( recurs to infinity.


How many moles of aluminum oxide can be made if 5.23 mol Al completely react?

Since the ratio of moles of Al to moles of Al2O3 is 4:2, if 5.23 mol Al completely reacts, 2.615 mol Al2O3 can be made.


How many moles of water are produced from 373 mol Al?

The balanced chemical equation for the reaction of aluminum (Al) with water (H2O) to produce aluminum oxide (Al2O3) and hydrogen gas (H2) is: 4 Al + 3 H2O → 2 Al2O3 + 3 H2. From the balanced equation, we can see that 3 moles of water are needed for every 4 moles of aluminum reacting. Therefore, if 373 mol of aluminum is reacting, then (373 mol Al) x (3 mol H2O / 4 mol Al) = 279.75 mol of water would be produced.


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