The reaction is:
4 Fe + 3 O2 = 2 Fe2O3
The answer is 319,38 g.
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.
Iron oxide is formed by the reaction of iron and oxygen in a 1:1 ratio by mass. Therefore, the 55 g of iron will react completely with 55 g of oxygen to form iron oxide.
Iron is typically produced through a chemical reduction reaction of iron ore (Fe2O3) with carbon in a blast furnace, resulting in the production of metallic iron. This reaction is a form of chemical reduction, where iron oxide is reduced to elemental iron by the carbon monoxide produced in the furnace.
The molecular mass of iron(III) chloride is 55.8 + 3(35.5) = 162.3 Amount of iron (III) chloride in a 80.5g pure sample = 80.5/162.3 = 0.496mol
One molecule of iron oxide (Fe₂O₃) contains two iron atoms.
Iron (III) oxide -> Iron + OxygenSteps to find the mass of iron obtained from 268g of iron (III) oxide:Step 1: Write the balanced equation2Fe2O3 -> 4Fe + 3O2Step 2: Identify key reactants and products2Fe2O3 -> 4FeStep 3: Write down the mole ratio2 moles : 4 molesStep 4: Change moles into gram formula mass2((2x56)+(3x16)) : 4(1x56)320g : 224gStep 5: Find out how many grams of iron will be produced for every gram of iron (III) oxide1 gram = 224/320 = 0.7gStep 6: Find out how many grams of iron will be produced for 268 grams of iron (III) oxide268 grams = (224/320)x268 = 187.6 gramsThe mass of iron produced from 268 grams of iron (III) oxide is 187.6 grams.
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.
Rusted iron is FeO, iron oxide. 63.8 grams FeO (1 mole FeO/71.85 grams)(6.022 X 1023/1 mole FeO) = 5.35 X 1023 atoms of iron oxide -------------------------------------------
Iron oxide is formed by the reaction of iron and oxygen in a 1:1 ratio by mass. Therefore, the 55 g of iron will react completely with 55 g of oxygen to form iron oxide.
26.20
155.2 g
i dont know help :C
The complete question: Lead (II) oxide reacts with ammonia forming solid lead nitrogen gas and liquid water. 1.)How many grams of ammonia are consumed in the reaction of 75.0g lead (II) oxide? 2.) If 56.4g of lead are produced how many grams of nitrogen are also formed?
Iron is typically produced through a chemical reduction reaction of iron ore (Fe2O3) with carbon in a blast furnace, resulting in the production of metallic iron. This reaction is a form of chemical reduction, where iron oxide is reduced to elemental iron by the carbon monoxide produced in the furnace.
The molecular mass of iron(III) chloride is 55.8 + 3(35.5) = 162.3 Amount of iron (III) chloride in a 80.5g pure sample = 80.5/162.3 = 0.496mol
1.000 kg (1000 grams/1 kg)(1000 mg/1 gram) = 1,000,000 mg/50 mg = 20,000 iron tablets produced
One molecule of iron oxide (Fe₂O₃) contains two iron atoms.