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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
What else can I help you with?
In the reaction of aluminum metal and oxygen gas aluminum oxide is formed How many moles of O2 will react with 2.0 moles Al?
In the reaction 4 moles of aluminum will react with 3 moles of oxygen to form 2 moles of aluminum oxide. Since we have 2.0 moles of aluminum, we would need (2.0 mol Al) x (3 mol O2 / 4 mol Al) = 1.5 moles of O2 to react with it.
How many grams of oxygen gas are required to react completely with 14.6 grams of solid sodium to form sodium oxide?
Balanced equation. 4Na + O2 ->2Na2O 14.6 grams Na (1 mole Na/22.99 grams)(1 mole O2/4 mole Na)(32.0 grams/1 mole O2) = 5.08 grams oxygen gas needed --------------------------------------------
Sodium will react with oxygen to form sodium oxide. Use the formula below to determine how many grams of sodium oxide will be formed if 46 grams of sodium are used in the reaction.?
62 grams a+
A gas mixture contains 26.3 grams of nitric oxide and 36.2 grams of oxygen gas What is the mole frraction of nitric oxide?
To find the mole fraction of nitric oxide, first calculate the moles of nitric oxide and oxygen gas separately by dividing the given mass by their respective molar masses. Then, find the total moles of both gases present in the mixture. Finally, divide the moles of nitric oxide by the total moles to calculate its mole fraction.
How many moles are in 28 grams of calcium oxide?
To find the number of moles in 28 grams of calcium oxide, we need to divide the given mass by the molar mass of calcium oxide. The molar mass of calcium oxide (CaO) is 56.08 g/mol. So, 28 grams of CaO is equal to 28 g / 56.08 g/mol = 0.5 moles of calcium oxide.
How many grams of aluminum oxide can be formed by the reaction of 1020g of aluminum with oxygen?
To determine the grams of aluminum oxide formed, we need to consider the balanced chemical equation for the reaction between aluminum and oxygen. The molar ratio between aluminum and aluminum oxide is 4:2. So, first calculate the moles of aluminum in 1020g, then use this to find the moles of aluminum oxide produced, and finally convert moles of aluminum oxide to grams.
What is 18.32 grams of aluminum and excess oxygen how many grams of aluminin oxiode are preduced?
Given the balanced chemical equation: 4Al + 3O2 → 2Al2O3, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. In this case, 18.32 grams of aluminum is equivalent to 0.684 moles. Using stoichiometry, we find that this would produce 0.456 grams of aluminum oxide.
How many moles of oxygen are produced when 26.5 mol of aluminum oxide are decomposed?
When aluminum oxide decomposes, it produces 2 moles of aluminum and 3 moles of oxygen for every mole of aluminum oxide. Therefore, for 26.5 moles of aluminum oxide decomposed, 3 * 26.5 = 79.5 moles of oxygen are produced.
What is the mole ratio of aluminum to oxygen?
The mole ratio of aluminum to oxygen in aluminum oxide (Al2O3) is 4:3, which means for every 4 moles of aluminum, there are 3 moles of oxygen.
What is the theoretical yield of aluminum oxide if 3.40mol of aluminum metal is exposed to 2.85mol of oxygen?
The balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide is: 4Al + 3O2 → 2Al2O3. From the balanced equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. Therefore, if 3.40 mol of aluminum and 2.85 mol of oxygen are reacted, the limiting reactant is oxygen. Thus, 3.40 mol of aluminum would theoretically produce 1.90 mol of aluminum oxide.
What is the theoretical yield of aluminum oxide if 3.20 mol of aluminum metal is exposed to 2.70 mol of oxygen?
The balanced chemical equation for the formation of aluminum oxide is: 4Al + 3O2 -> 2Al2O3. This means that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. Using the given moles of aluminum and oxygen, we can determine the limiting reactant and the theoretical yield of aluminum oxide. In this case, the oxygen is the limiting reactant, which means it will run out first. Therefore, all of the 2.70 mol of oxygen will react with aluminum to produce aluminum oxide. Since the ratio of aluminum to aluminum oxide is 4:2, the 2.70 mol of oxygen will react with (2/3)*2.70 mol of aluminum to produce aluminum oxide, which is approximately 1.80 mol.
What is the stoichiometry behind making aluminum oxide?
The stoichiometry for making aluminum oxide (Al2O3) involves the reaction between aluminum metal and oxygen gas. The balanced chemical equation is 4Al + 3O2 -> 2Al2O3, which means that 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide.
How many grams of Aluminium oxide will be produced from the complete reaction of 1.87 L of oxygen gas at 793.0 mm Hg and 28.0 Celsius degree?
To find the grams of aluminum oxide produced, we need to first calculate the moles of oxygen gas using the ideal gas law. Then, we use the balanced chemical equation to determine the moles of aluminum oxide produced. Finally, we convert moles to grams using the molar mass of aluminum oxide. Be sure to adjust the conditions of the gases to standard temperature and pressure for accurate calculations.
How many moles of aluminum oxide can be made if 4.0 moles Al completely react?
The balanced chemical equation for the reaction between aluminum (Al) and oxygen (O₂) to form aluminum oxide (Al₂O₃) is: [ 4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3 ] According to the balanced equation, 4 moles of aluminum (Al) produce 2 moles of aluminum oxide (Al₂O₃). Therefore, if 4.0 moles of aluminum completely react, it will produce ( \frac{2}{4} \times 4.0 ) moles of aluminum oxide. Calculate that to find the answer.
What is the theoretical yield of aluminum oxide if 2.40 mole of aluminum metal is exposed to 2.10 mol of oxygen?
To determine the theoretical yield of aluminum oxide, we first need to write a balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide. The balanced equation is 4Al + 3O2 -> 2Al2O3. From the equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. Therefore, if 2.40 moles of aluminum is exposed to 2.10 moles of oxygen, the limiting reactant is oxygen. Using stoichiometry, we can calculate the theoretical yield of aluminum oxide, which is 1.60 moles.
What does the chemical equation 2AL2O3 mean?
The chemical equation 2Al2O3 represents the reaction of two aluminum atoms with three oxygen atoms to form two molecules of aluminum oxide. This equation shows that for every two moles of aluminum, three moles of oxygen are required to form two moles of aluminum oxide.
How many moles of aluminum oxide are present in a sample having a mass of 178.43g?
For this you need the atomic (molecular) mass of Al2O3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. Al2O3= 102 grams408 grams Al / (102 grams) = 4.00 moles Al
