To form iron(III) oxide, the chemical equation is:
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
From the equation, it can be seen that 3 moles of oxygen (O2) are needed to react with 4 moles of iron (Fe) to produce 2 moles of iron(III) oxide (Fe2O3). The molar mass of oxygen (O2) is approximately 32 g/mol. Therefore, you would need 96 grams of oxygen to react with the iron needed to form iron(III) oxide.
To determine the amount of iron needed to react with 40 grams of iron(III) oxide, you should use the stoichiometry of the reaction. Calculate the molar mass of iron(III) oxide (Fe2O3) and determine the molar ratio between iron and iron(III) oxide in the balanced chemical equation. From there, you can calculate the amount of iron needed to fully react with 40 grams of iron(III) oxide.
To find the mass of oxygen needed to produce 13 grams of the oxide, we first find the molar ratio between P and O in the product. Then, use this ratio to calculate the mass of oxygen needed. Since 71g of the oxide contains 31g of phosphorus, we can calculate the mass of oxygen needed for 13g of the oxide by setting up a proportion.
You would need 5.7 grams of mercury II oxide to produce 1.56 L of oxygen gas according to the following reaction at those conditions.
To find the grams of uranium oxide formed, we need to determine the molar mass of uranium and oxygen, calculate the moles of each element present, and finally the moles of uranium oxide formed. Then, we convert moles to grams using the molar mass of uranium oxide. The final answer for the grams of uranium oxide formed depends on the stoichiometry of the reaction.
To determine the amount of oxygen, we first find the amount of magnesium by subtracting the given 20.0 grams of magnesium oxide from the total. Given that the molar mass of magnesium oxide is 40.3 g/mol and that of magnesium is 24.3 g/mol, we calculate the amount of oxygen by adjusting accordingly. This process gives us the weight ratio of magnesium oxide to oxygen.
To determine the amount of iron needed to react with 40 grams of iron(III) oxide, you should use the stoichiometry of the reaction. Calculate the molar mass of iron(III) oxide (Fe2O3) and determine the molar ratio between iron and iron(III) oxide in the balanced chemical equation. From there, you can calculate the amount of iron needed to fully react with 40 grams of iron(III) oxide.
To find the mass of oxygen needed to produce 13 grams of the oxide, we first find the molar ratio between P and O in the product. Then, use this ratio to calculate the mass of oxygen needed. Since 71g of the oxide contains 31g of phosphorus, we can calculate the mass of oxygen needed for 13g of the oxide by setting up a proportion.
2
You would need 5.7 grams of mercury II oxide to produce 1.56 L of oxygen gas according to the following reaction at those conditions.
To find the grams of uranium oxide formed, we need to determine the molar mass of uranium and oxygen, calculate the moles of each element present, and finally the moles of uranium oxide formed. Then, we convert moles to grams using the molar mass of uranium oxide. The final answer for the grams of uranium oxide formed depends on the stoichiometry of the reaction.
If 3 grams of magnesium are used to form 4 grams of magnesium oxide, then 1 gram of oxygen is used in the reaction. This means 1 gram of oxygen remains unused.
To determine the amount of oxygen, we first find the amount of magnesium by subtracting the given 20.0 grams of magnesium oxide from the total. Given that the molar mass of magnesium oxide is 40.3 g/mol and that of magnesium is 24.3 g/mol, we calculate the amount of oxygen by adjusting accordingly. This process gives us the weight ratio of magnesium oxide to oxygen.
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 --------------------------------------------
same thing as oxygen no subscripts needed
To find the amount of oxygen needed to produce 95.6 g of aluminum oxide (Al2O3), first calculate the molar mass of Al2O3 (101.96 g/mol). Then, set up a ratio using the molar mass ratio of oxygen to Al2O3 (3:2). Calculate the amount of oxygen needed using the given mass of Al2O3 and the molar ratio.
For every 40 grams of calcium (Ca), 32 grams of oxygen (O) will be needed to react. This is based on the chemical formula for calcium oxide (CaO), where one calcium atom reacts with one oxygen atom to form one molecule of CaO.
Nitrogen and Oxygen