80.0 g of iodine (v) oxide reacts with 28.0 g of carbon monoxide determine the mass of iodine

Well, let's solve this problem with a touch of humor!

To begin, let's write the balanced chemical equation for the reaction:

I2O5 + 5 CO → I2 + 5 CO2

Now, let's find the molar masses of iodine (V) oxide and carbon monoxide. The molar mass of I2O5 is approximately 319.81 g/mol, while the molar mass of CO is around 28.01 g/mol.

Given that you have 80.0 g of I2O5 and 28.0 g of CO, we need to determine which one is the limiting reagent, which means the reactant that will run out first and determine the amount of product formed.

To do this, we'll use the molar masses and convert the masses to moles.

The number of moles of I2O5 can be calculated as:

80.0 g / 319.81 g/mol = 0.250 mol

Similarly, the number of moles of CO is:

28.0 g / 28.01 g/mol = 1.00 mol

According to the stoichiometry of the balanced equation, the ratio of I2O5 to CO is 1:5. Therefore, to completely react with the given amount of I2O5, we would need 5 times the number of moles of CO, which is 5.00 mol.

But, since we only have 1.00 mol of CO, it is the limiting reagent.

Now, let's determine the mass of iodine (I2) that will be produced.

Using the ratio from the balanced equation, 1 mol of I2O5 produces 1 mol of I2, so 0.250 mol of I2O5 will produce 0.250 mol of I2.

Finally, let's convert the number of moles to grams using the molar mass of iodine, which is around 253.80 g/mol.

0.250 mol * 253.80 g/mol = 63.45 g

Therefore, the mass of iodine (I2) produced in this reaction is approximately 63.45 grams.

To determine the mass of iodine, we need to calculate the stoichiometry of the reaction between iodine (V) oxide (I2O5) and carbon monoxide (CO).

The balanced chemical equation for the reaction can be written as follows:
I2O5 + 5CO -> 5CO2 + 2I2

From the balanced equation, we can see that the molar ratio between I2O5 and I2 is 2:1. This means that for every mole of I2O5, we will produce 2 moles of I2.

First, let's calculate the number of moles of I2O5 and CO:

Number of moles of I2O5 = Mass of I2O5 / Molar mass of I2O5
Number of moles of CO = Mass of CO / Molar mass of CO

The molar masses of I2O5 and CO can be calculated using the periodic table:

Molar mass of I2O5 = 2(I) + 5(O) = 2(126.90 g/mol) + 5(16.00 g/mol) = 333.80 g/mol
Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol

Substituting the given values, we can calculate the number of moles:

Number of moles of I2O5 = 80.0 g / 333.80 g/mol ≈ 0.240 moles
Number of moles of CO = 28.0 g / 28.01 g/mol ≈ 0.999 moles

Since the molar ratio between I2O5 and I2 is 2:1, the number of moles of iodine (I2) produced will be half the number of moles of I2O5:

Number of moles of I2 = 0.240 moles / 2 ≈ 0.120 moles

Finally, to find the mass of iodine (I2), we can use the molar mass of I2:

Mass of I2 = Number of moles of I2 * Molar mass of I2
Mass of I2 = 0.120 moles * 253.80 g/mol = 30.45 g (rounded to two decimal places)

Therefore, the mass of iodine produced in the reaction is approximately 30.45 grams.

To determine the mass of iodine, we need to calculate the amount of iodine consumed in the reaction using the given information about the reactants.

1. Convert the given masses of iodine (V) oxide and carbon monoxide into moles using their respective molar masses:

- Iodine (V) oxide (I2O5): molar mass = 214 g/mol
- Carbon monoxide (CO): molar mass = 28 g/mol

Moles of I2O5 = mass / molar mass = 80.0 g / 214 g/mol = 0.373 moles
Moles of CO = mass / molar mass = 28.0 g / 28 g/mol = 1 mole

2. Use the balanced chemical equation for the reaction between iodine (V) oxide and carbon monoxide to determine the stoichiometry:

I2O5 + 5CO -> I2 + 5CO2

According to the balanced equation, 1 mole of I2O5 reacts with 1 mole of I2.

3. Since the stoichiometry of the reaction is 1:1, the moles of iodine (I2) produced will be equal to the moles of iodine (V) oxide consumed.

Moles of I2 = Moles of I2O5 = 0.373 moles

4. Finally, calculate the mass of iodine (I2) using its molar mass:

Molar mass of iodine (I2) = 253.8 g/mol

Mass of iodine (I2) = moles x molar mass = 0.373 moles x 253.8 g/mol = 94.65 g

Therefore, the mass of iodine produced in the reaction is approximately 94.65 grams.

I2O5+CO>> I2 + CO2

balance...

I2O5 + 5CO>>I2 + 5CO2

so the next is to see how many moles you have
28gCO/28= 1mol CO
80/334= .24 moles.

Since you need 5 mole CO per mole of I2O5, then I2O5 is limiting, and you have CO in excess.

so you will get .24mol I2, convert that to grams.