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Identify the number of moles of excess reagent left over when 4.821 moles of ironIII oxide are reacted with 5.591 moles of aluminum in the thermite reaction?

The balanced equation for the thermite reaction is: 2Al + Fe2O3 -> Al2O3 + 2Fe. This means that 2 moles of aluminum react with 1 mole of ironIII oxide. Therefore, if 5.591 moles of aluminum are used, 2.7955 moles of ironIII oxide will react. Since 4.821 moles of ironIII oxide are provided, the excess reagent is aluminum. The amount of excess aluminum left over is 5.591 moles - 2.7955 moles = 2.7955 moles.


Consider a reaction between 5.39 moles of ironIII oxide and excess aluminum in the thermite reaction If 526 g of liquid iron were recovered after the reaction was complete what was the percent yield o?

87.4%


Consider a reaction between 8.63 moles of ironIII oxide and excess aluminum in the thermite reaction If 856 g of liquid iron were recovered after the reaction was complete what was the percent yield o?

88.8%


Aluminum will react with ironIII oxide to form aluminum oxide and iron according to this reaction?

The reaction you are referring to is a displacement reaction in which aluminum replaces iron in iron(III) oxide to form aluminum oxide and iron. The balanced equation for the reaction is: 2Al + Fe2O3 -> Al2O3 + 2Fe


What is the reaction between aluminum and ironIII oxide that can generate temperatures approaching 30000 degrees C and is used in welding metals?

It is called the thermite reaction, and it's expressed this way: Fe2O3 + 2Al -> 2Fe + Al2O3 + Heat A link can be found below. It's to the Wikipedia post, and you'll get more information there.


What is the maximum mass of iron metal if 50.0 grams of ironIII oxide is mixed with 50.0 grams of aluminum?

Fe2O3 + 2Al --> 2Fe + Al2O3Before:50.0g + 50.0g > 0.0g + (not important)159.69(g/mol) + 26.98(g/mol)In mol (before reaction):+0.3131 mol + 1.853 mol (excess)Reaction (used reactant > formed Fe):-0.3131 mol - 0.6262 mol > + 0.6262 mol FeRemaining (= before - used):0.0 mol Fe2O3 + 1.227 mol Al > 0.6262 mol Fe, this should be multiplied by the molar mass of Fe to get mass in grams: 0.6262 (molFe) * 55.85 (g/molFe) = 34.97 = 35.0 g Fe


How many moles of ironIII hydroxide are needed to react with excess nitric acid to produce 63.8 g ironIII nitrate?

Balanced equation. Fe(OH)3 + 3HNO3 --> Fe(NO3)3 + 3H2O 63.8 grams Fe(NO3)3 (1 mole Fe(NO3)3/241.88 grams)(1 mole Fe(OH)3/1 mole Fe(NO3)3 = 0.264 moles iron III hydroxide needed ==========================


Is ironIII bromide?

FeBr3


What is the compound formula of ironIII oxide?

Fe2O3


What is the chemical formula for IronIII Bromide?

Formula: FeBr3


What is the oxidation number of ironIII?

Iron(iii) ion = +3


Is ironIII a cation?

Yes it is and is also called ferric cation