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In the formula Fe2O3, there are two iron (Fe) atoms and three oxygen (O) atoms. The molar mass of iron is approximately 56 amu, so the mass contributed by iron is 2 × 56 amu = 112 amu. Therefore, the mass of oxygen in Fe2O3 is the total formula mass (160 amu) minus the mass of iron (112 amu), which equals 48 amu. Thus, the mass of oxygen in Fe2O3 is 48 amu.
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When iron reacts with oxygen it forms iron oxide or rust. 4 Fe plus 3 O2 2 Fe2O3 If 112 g of iron (Fe) combines with 24 g of oxygen (O2) how much iron oxide is formed?
To determine the amount of iron oxide (Fe2O3) formed, we first need to identify the limiting reactant. The molar mass of Fe is approximately 56 g/mol and for O2, it is about 32 g/mol. Calculating the moles, 112 g of Fe gives 2 moles (112 g / 56 g/mol), and 24 g of O2 gives 0.75 moles (24 g / 32 g/mol). The balanced reaction shows that 4 moles of Fe react with 3 moles of O2; thus, O2 is the limiting reactant. According to the stoichiometry, 0.75 moles of O2 will produce 1.0 mole of Fe2O3, which corresponds to approximately 160 g (1 mole of Fe2O3 is 160 g). Therefore, 160 g of iron oxide is formed.
When you have 46 grams of sodium react with 160 grams of oxygen how many moles of sodium oxide are obtained?
The equation for the reaction is 4 Na + O2 -> 2 Na2O. This shows that, for complete reaction, one mole of oxygen is required for each four gram atomic masses of sodium. The gram atomic mass of sodium is 22.9898; therefore, 46 grams of sodium constitutes 2.00 moles of sodium, to more than the justified number of significant digits. The gram molecular mass of diatomic oxygen is 31.9988; therefore 160 grams of oxygen constitutes 5.000 moles of diatomic oxygen, to more than the justified number of significant digits. This is well over the minimum amount of oxygen required for complete reaction of all the sodium present. Each two gram atomic masses of sodium produces one gram formula mass of sodium oxide; therefore, the number of gram formula masses of sodium oxide produced is 1.00, to at least the justified number of significant digits.
How many moles of Fe are present in 30 grams in rust?
This amount may be different because rust is not a clearly definite compound.
How many moles there are in 160g of argon atoms?
To find the number of moles of argon in 160 grams, you can use the molar mass of argon, which is approximately 40 g/mol. Using the formula: [ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} ] we can calculate: [ \text{Number of moles} = \frac{160 \text{ g}}{40 \text{ g/mol}} = 4 \text{ moles} ] Thus, there are 4 moles of argon in 160 grams.
How long is 160 mm in inches?
160mm is 6.29921 inches. Direct Conversion Formula 160 mm* 1 in 25.4 mm = 6.299212598 in
What is the chemical formula for Fe2O3?
That is the chemical formula. The name would be iron (III) oxide, or ferric oxide in the old system.
If the substance has a molar mass of 160 plus 5 grams per mole what is its molecular formula?
To find the molecular formula, you need the empirical formula and molar mass. If the molar mass is 160 plus 5 grams per mole, the molecular formula cannot be determined without additional information about the empirical formula's molar mass relationship.
Consider the following balanced equation Fe203 plus 3H2 gives 2Fe plus 3H2O What mass of hydrogen would be required to convert 160g of iron III oxide into iron?
6g hydrogen would be required for 160g ferric oxide in this reaction. The relative atomic weights of the elements are: Hydrogen - 1 Oxygen - 16 Iron - 56 giving the relative atomic weights of the compounds (on the left of the equation): Fe2O3 = 56×2 + 16×3 = 160 3H2 = 3×(1×2) = 6 So for every 160 units of mass of iron III oxide there will be 6 units of mass of hydrogen required. → for 160g of iron III oxide ÷ 160 × 6 = 6 g of hydrogen.
Why is 160 grams of oxygen mass would be produced?
This question seems to be about the reactant side of a chemical equation. To calculate the mass of oxygen produced, you need to know the stoichiometry of the reaction. Without that information, it is not possible to determine why 160 grams of oxygen would be produced.
When iron reacts with oxygen it forms iron oxide or rust. 4 Fe plus 3 O2 2 Fe2O3 If 112 g of iron (Fe) combines with 24 g of oxygen (O2) how much iron oxide is formed?
To determine the amount of iron oxide (Fe2O3) formed, we first need to identify the limiting reactant. The molar mass of Fe is approximately 56 g/mol and for O2, it is about 32 g/mol. Calculating the moles, 112 g of Fe gives 2 moles (112 g / 56 g/mol), and 24 g of O2 gives 0.75 moles (24 g / 32 g/mol). The balanced reaction shows that 4 moles of Fe react with 3 moles of O2; thus, O2 is the limiting reactant. According to the stoichiometry, 0.75 moles of O2 will produce 1.0 mole of Fe2O3, which corresponds to approximately 160 g (1 mole of Fe2O3 is 160 g). Therefore, 160 g of iron oxide is formed.
How many grams of c are required to react with 16.5g of fe2o3?
The balanced reaction equation is 2Fe2O3 + 3C = 4Fe + 3CO2 Note the molar ratios are 2:3::4:2 Next calculate the acutal moles of Fe2O3 First using the Periodic Table Calculate the Mr ( Relative molecular mass) of Fe2O3 Fe x 2 = 56 x 2 = 112 O x 3 = 16 x 3 = 48 112 + 48 = 160 mol(Fe2O3) = 16.5 / 160 = 0.103125 Equivlanet to ;2; moles. mol(C) = 0.103125 x 3/2 = 0.1546875 equivalent to '3 moles. Mass(C) = 0.154687 x 12 = 1.85625 g (The Answer!!!!)
How many grams of fe2o3 are required to completely react with 84 grams of co?
160...cant quite grasp HOW though
The rock in a particular iron ore deposit contains 78 percent Fe2O3 by mass how many kilograms of the rock must be processed to obtain 1000kg of iron?
Good one. Fe2O3 is 160g/mol, so %comp of Fe is 112/160=0.7 or 70%. 1x106g Fe/70=x/30, so x=428,571g O2, so mass of Fe2O3 is 1,428,571g. 1,428,571/78=x/22, so x=402,930g of other rock, so mass of ore deposit = 1,428,571+402,930=1,831,501g=1,831.5kg.
When you have 46 grams of sodium react with 160 grams of oxygen how many moles of sodium oxide are obtained?
The equation for the reaction is 4 Na + O2 -> 2 Na2O. This shows that, for complete reaction, one mole of oxygen is required for each four gram atomic masses of sodium. The gram atomic mass of sodium is 22.9898; therefore, 46 grams of sodium constitutes 2.00 moles of sodium, to more than the justified number of significant digits. The gram molecular mass of diatomic oxygen is 31.9988; therefore 160 grams of oxygen constitutes 5.000 moles of diatomic oxygen, to more than the justified number of significant digits. This is well over the minimum amount of oxygen required for complete reaction of all the sodium present. Each two gram atomic masses of sodium produces one gram formula mass of sodium oxide; therefore, the number of gram formula masses of sodium oxide produced is 1.00, to at least the justified number of significant digits.
What is the total mass of 222Rn remaining in an original 160-milligram sample of 222Rn after 19.1 days?
To find the remaining mass of a radioactive isotope after a certain time, you can use the radioactive decay formula: [M_{\text{final}} = M_{\text{initial}} \times \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}] Given that the half-life of (^{222}\text{Rn}) is 3.8 days, and the initial mass is 160 milligrams, you can substitute these values into the formula to find the final mass.
What is the volume of 160 kilograms of methylated spirit?
V (VOLUME) = M (MASS) / D (DENSITY) All you have to do is to find the density of methylated spirit and use the formula
How many moles of Fe are present in 30 grams in rust?
This amount may be different because rust is not a clearly definite compound.
