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The formula for aluminum oxide is Al_2O_3. The correct answer is Al_2O_3. Let us see how we got the answer; Look at the electronic arrangement of Al and O atoms. Al ( Z= 13) has 13 electrons with following electronic configuration. 1s^22s^22p^63s^23p^1 It loses three electron in its 3s and 3p subshell to achieve stability and forms ion Al^(3+). Al^(3+) = 1s^22s^22p^6 O ( Z=8) on the other hand ...
You can determine the empirical formula for this compound (or any compound, for that matter) by finding the mole ratio between oxygen and aluminium. FIrst, you calculate the number of moles that reacted for both elements - it's a little easier to use monoatomic oxygen, not diatomic oxygen "4.151 g" * "1 mole aluminium"/"26.98 g" = "0.1539 moles " Al, and "3.692 g" * ("1 mole monoatomic oxygen ...
Aluminum ions have a 3+ charge ("Al"^(3+) and oxide ions have a 2- charge ("O"^(2-)). Use the crossover rule to determine the formula, in which the size of the charge on one ion becomes the subscript on the other ion.
For the example Aluminum Oxide #Al_2O_3# Aluminum has an oxidation state of +3 or Al+3 Oxygen has an oxidation state of -2 or #O^−2# The common multiple of 2 and 3 is 6.? We will need 2 aluminum atoms to get a +6 charge and 3 oxygen atoms to get a -6 charge. When the charges are equal and opposite the atoms will bond as #Al_2O_3#.
4"Al" + 3"O"_2rarr2"Al"_2"O"_3 "Al" + "O"_2rarr"Al"_2"O"_3 This is the unbalanced equation described. Achieve a common 6 oxygen atoms on either side of the equation. "Al" + 3"O"_2rarr2"Al"_2"O"_3 Now, since there are 1 and 4 aluminum atoms on each side respectively, change the coefficient on the reactants side to 4. Balanced form: 4"Al" + 3"O"_2rarr2"Al"_2"O"_3
Al_2(SO_4)_3 -> Al_2O_3 + 3SO_3 Putting 3 as cooefficient for sulfur trioxide will balance the chemical equation. Check the number of atoms of each element by having an inventory. Reactant Side: Al=2xx1=2 S=3xx1xx1=3 O=3xx4xx1=12 Product Side: Al=2xx1=2 S=1xx3=3 O=3xx1 + 3xx3 = 12
The empirical formula is the simplest formula of a compound. The actual formula is an integral multiple of the empirical formula. Let's assume that the molecular mass turned out to be about 180 u.
This is a synthesis reaction. We're given that aluminum ("Al") can react with oxygen gas ("O"_2; don't forget it's diatomic!) to produce aluminum oxide ("Al"_2"O"_3). The balanced chemical equation for this reaction is ul(4"Al"(s) + 3"O"_2(g) rarr 2"Al"_2"O"_3(s) A reaction in which two or more reactants combine to form only one product is called a synthesis reaction. Here, there are two ...
You calculate the molar ratios of each element in the oxide. EXAMPLE. When a 2.50 g sample of copper is heated, it forms 3.13 g of an oxide. What is its empirical formula. Solution. Step 1. Determine the masses. Mass of Cu = 2.50 g. Mass of O = (3.13 – 2.50) g = 0.63 g. Step 2. Determine the moles
Single displacement reaction This is because aluminium molecule does not give anything to iron but iron gives its oxide atom to alimunium. So only one or single molecule that is iron oxide displaces its oxide atom so this a "Single displacement reaction"