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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 --------------------------------------------
First determine the balanced chemical reaction:4Al + 3O2 ---> 2Al2O3Next figure out the number of moles of oxygen you start with by dividing 192 by the molecular weight of oxygen (16 g/mol). Then use the coefficients from the reaction to determine how many moles of Al would need to react with the oxygen to convert all the oxygen and aluminum to aluminum oxide.192 g * (1 mol O / 16 g O) * (4 moles of Al/6 moles of O) = 8 mol of AlFinally multiply by the molecular weight of Aluminum to figure out the mass:8 mol Al * (27 g/mol) = 216 g of 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.
To calculate the number of grams of oxygen needed to react with 6.78 grams of ammonia, we first write out the balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O). Then we use the stoichiometry of the equation to find the molar ratio between ammonia and oxygen. Finally, we convert the mass of ammonia to moles and then use the molar ratio to find the mass of oxygen needed.
The balanced chemical equation for the reaction of hydrogen and oxygen to form water is 2H2 + O2 -> 2H2O. Based on the equation, for every 2 grams of hydrogen, 64 grams of oxygen are needed to form 36 grams of water. Thus, if 8 grams of hydrogen react completely with 64 grams of oxygen, the total mass of water formed would be 36 grams.
Three atoms of oxygen are required to react with each two atoms of aluminum to form the most common product of reaction between oxygen and aluminum. Therefore, 0.75 mole of oxygen atoms will be required to react with 0.5 mole of aluminum atoms. The atomic weight of oxygen is 15.999; therefore, the mass will be (0.75)(15.999) = 12 grams of oxygen, to the maximum possibly justified number of significant digits.
The answer is 152 g oxygen.
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.
The balanced equation for the reaction between ammonia (NH3) and oxygen (O2) is 4NH3 + 5O2 → 4NO + 6H2O. To find the grams of oxygen needed to react with 23.9 grams of ammonia, you need to calculate the molar ratio between ammonia and oxygen using the balanced equation. Once you find the molar ratio, you can calculate the grams of oxygen required.
266,86 g aluminium chloride are obtained.
The balanced chemical equation for the reaction is: 4Fe + 3O2 -> 2Fe2O3 From the equation, it can be seen that 3 moles of O2 are required to react with 4 moles of Fe. Therefore, to determine the grams of O2 required to react with 100 g Fe, you would need to use stoichiometry to find the answer.
To determine the amount of gas needed to react with 348.5 grams of oxygen, you need to know the balanced chemical equation of the reaction. Then, use the stoichiometry of the reaction to calculate the amount of gas required based on the molar ratio between the gas and oxygen in the reaction.
When you mix aluminum and oxygen, you get aluminum oxide. If you mix iron with aluminum oxide, the aluminum will react with the iron oxide, forming a thermite reaction that produces molten iron and aluminum oxide slag.
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.
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 --------------------------------------------
First determine the balanced chemical reaction:4Al + 3O2 ---> 2Al2O3Next figure out the number of moles of oxygen you start with by dividing 192 by the molecular weight of oxygen (16 g/mol). Then use the coefficients from the reaction to determine how many moles of Al would need to react with the oxygen to convert all the oxygen and aluminum to aluminum oxide.192 g * (1 mol O / 16 g O) * (4 moles of Al/6 moles of O) = 8 mol of AlFinally multiply by the molecular weight of Aluminum to figure out the mass:8 mol Al * (27 g/mol) = 216 g of Al
To find the grams of bromine that react with 15.0g of aluminum, we need to first calculate the molar mass of aluminum and aluminum bromide. Then, find the moles of aluminum reacted using its mass and molar mass. Using the mole ratio from the balanced chemical equation, we can find the moles of bromine that react. Finally, convert the moles of bromine to grams using its molar mass.