Assigned Reaction IRON(lll) ➕OXYGEN
By Abdel Diab
IUPAC Name for each Reactant AND Product
Type of Reaction
4Fe ➕ 3O2 ➡➡➡ 2Fe2O3
We cris cross the charges of the two elements
First, we need to balance the oxygen on each side which would require from us to add a coefficient 2 to the compound so that the O3 and O2 balance. That made 2Fe2 which basically means 4 Fe's so we added 4 to the Fe on the left side
The Uses of Iron(lll)oxide in Real World
- The reddish color of the needles in canyonlands national park in Utah is due to the high amounts of Fe2O3
- Gives it's rust color to iron ores and various clay minerals that contain Fe2O3.
- It can be used to make thermite which can be used for welding.
- Can be used as metal polish to keep fresh metal underneath it and can also be chipped off easily
- Used as a pigment for paints and cosmetics
Molar Mass for the Reactants and Products
Oxygen(O2) = 31.998 grams
Iron(lll)oxide = 159.687 grams
In calculating the molar mass, you must refer to the periodic table and look at the molar mass of each individual element and multiply it by the subscript, then add the sum for that one compound and move on to the next compound. Disregard the coefficient when you calculate the molar mass. Your reactants added up should be equal to your products added up.
Mole to Mole Conversion
To calculate mole to mole conversion, you must start with your given number of moles of compound "A" from the question and place that into the upper left box. In the lower right box, you must put the coefficient (moles) of compound "A" from the balanced equation. Above that, you put the coefficient (moles) of compound "B". Compound B is what the question asks you for. You multiply the top row, and then you divide by the bottom row.
Mass to Mass conversion
To calculate mass to mass conversion, you must start with the given mass (grams) of compound "A" and place that in the upper left box. Diagonally right on the bottom, you put the molar mass (grams) of that compound. This is calculated from the periodic table. Above that, you need to write the number of moles of compound "A" which is always one. Diagonally right on the bottom of that, you need to put the coefficient (moles) of compound "A", and above that, put the coefficient (moles) of compound "B". Next, in the lower right box you need to put the number of moles of compound "B" which is always one, and above that, the molar mass (grams) of that compound. This is also calculated from the periodic table. Multiply the top, and then divide by the bottom.
The steps to solve mass to Mass conversion
- Start off with the given which is 12.1 g of Iron(lll)
- Use the molar mass of Iron(lll) And put it at the bottom to cross out iron and 1 mole (always) of Iron(lll) is going to be at the top
- Always use the coefficient from the equation and put the element that you used first at the bottom and the second element/compound on top
- Use the molar mass of the other compound/element and put it on top and on the bottom is always going to be one because it's how many grams in one mole.
- Multiply the top and divide the bottom
Limiting and Excess Reactant
The limiting reactant is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reactant since the reaction cannot proceed further without it.
For this problem, Iron(lll) was the limiting reactant because the result was smaller and Oxygen was the excess because the result was bigger
The theoretical yield is the maximum amount of product that could possibly be produced in a given reaction. This calculation assumes that only one reaction occurs and that the limiting reactant reacts completely. The amount is limited by the limiting reactan.
The theoretical yield is the result of the limiting reactant mass to mass conversion equation.
In the picture above, the limiting reactant was Iron(lll) which means the theoretical yield (product) is 17.59 grams
The percent yield is the actual yield ➗ the theoretical yield.
In calculating percent yield, the numerator in the fraction will be given to you in the question. It it the actual yield. The denominator is the theoretical yield, and you get this from finding the limiting reactant. After dividing the numerator from the denominator, you multiply your answer by 100. The final answer is your percent yieldThis is the way of doing it following the example above :