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How many grams of iron will be needed to react completely with an excess of oxygen to form 40 grams of iron3 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.
A 55 g sample of iron reacts with 24 g of oxygen to form how many grams 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.
How many elements in iron oxide?
Iron oxide has two main chemical compounds: iron(II) oxide (FeO) and iron(III) oxide (Fe2O3). Both compounds consist of two elements - iron and oxygen.
How many grams of oxygen needed to get iron3 oxide?
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.
How many molecules of iron are there in iron oxide?
One molecule of iron oxide (Fe2O3) contains two iron atoms per molecule. So, for every molecule of iron oxide, there are two iron atoms, which means one iron oxide molecule contains two iron atoms.
How many grams of iron will be needed to react completely with an excess of oxygen to form 40 grams of iron3 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.
A 55 g sample of iron reacts with 24 g of oxygen to form how many grams 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.
How many atoms of iron rusted that weight 63.8 grams?
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 -------------------------------------------
Sodium oxide and water react to produce sodium hydroxide. How many grams of sodium hydroxide is produced from 120.0g of sodium oxide?
155.2 g
How many grams of water will be produced if 32.0 g of nitrous oxide are also produced in the reaction?
To find the mass of water produced, we first need to determine the molar ratio of water to nitrous oxide in the reaction. Once we have the mole ratio, we can use it to calculate the moles of water produced from the moles of nitrous oxide. Finally, we can convert the moles of water to grams using the molar mass of water.
How many grams of iron can be obtained from a 268g sample of iron 3 oxide?
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.
How many grams of ammonia are consumed in the reaction of 75.0g lead (II) oxide?
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?
How many elements in iron oxide?
Iron oxide has two main chemical compounds: iron(II) oxide (FeO) and iron(III) oxide (Fe2O3). Both compounds consist of two elements - iron and oxygen.
How many grams of oxygen needed to get iron3 oxide?
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.
How many molecules of iron are there in iron oxide?
One molecule of iron oxide (Fe2O3) contains two iron atoms per molecule. So, for every molecule of iron oxide, there are two iron atoms, which means one iron oxide molecule contains two iron atoms.
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 grams of iron must be added to 16 grams of oxygen to equal 44 grams of iron 3 oxide?
30.8g of iron (approximately) reacted with 13.2g of oxygen will yield 44g of iron(III) oxide (Fe2O3) with 2.8g of oxygen left unreacted. This assumes atomic mass numbers of 56 and 16 respectively for iron and oxygen. The actual mass number of iron is 55.847 and oxygen 15.9994 making the figures 30.775g of iron and 13.225g of oxygen with 2.775g of unreacted oxygen. Of course this is an exothermic reaction so will there be a tiny tiny loss of mass in the system as it is converted to heat energy, according to E=MC^2? I'll let you work that one out...
