How many grams of CO are needed to react with an excess of Fe2O3 to produce 209.7gFe? Fe2O3(s) + 3CO(g) => 3CO2(g) + 2Fe(s)

Moles Fe = 209.7 g / 55.847 = 3.75

the ratio between CO and Fe is 3 : 2
Moles CO needed = 3.75 x 3 / 2 = 5.63
Mass CO = 5.63 mol x 28.01 g/mol =157.7 g

is this right?

Convert 209.7 g Fe to mols. molsl = grams/molar mass

Using the coefficients in the balanced equation, convert mols Fe to mols CO.
Now convert mols CO to grams. g = mols x molar mass.

I'm not an expert on this subject, but I think that I understand it. The numbers that seem to come out of nowhere are the atomic masses of the various elements and compounds, and the calculations for those are all correct.

Why did the iron and carbon monoxide go on a road trip together? Because they wanted to COllide and form a Fe-rrific bond! But let's calculate how many grams of CO are needed to make this Fe-rrific road trip happen.

First, we need to figure out the molar mass of Fe, which is 55.85 g/mol. Since we know that we want to produce 209.7g of Fe, we can calculate the number of moles of Fe using the equation:

moles of Fe = mass of Fe / molar mass of Fe

moles of Fe = 209.7g / 55.85 g/mol

Next, we can use the balanced equation to determine the ratio of moles of CO to moles of Fe. According to the equation, 3 moles of CO react with 2 moles of Fe. Therefore, the ratio is 3:2.

Now, we can calculate the moles of CO needed using the ratio of moles of CO to moles of Fe:

moles of CO = (moles of Fe) x (3 moles of CO / 2 moles of Fe)

Finally, we can convert the moles of CO to grams using the molar mass of CO, which is 28.01 g/mol:

grams of CO = (moles of CO) x (molar mass of CO)

And there you have it! Calculate these steps, and you'll find out how many grams of CO are needed for this Fe-rrific reaction to occur.

To determine the number of grams of CO required to react with an excess of Fe2O3, we need to use stoichiometry.

The balanced chemical equation given is:
Fe2O3(s) + 3CO(g) → 3CO2(g) + 2Fe(s)

From the equation, we can see that the ratio between Fe2O3 and CO is 1:3. This means that for every 1 mole of Fe2O3, we need 3 moles of CO.

To find the number of moles of Fe2O3, we can calculate as follows:
moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3

Given that the mass of Fe produced is 209.7g, we can assume the mass of Fe2O3 used is also 209.7g. This is because Fe2O3 and Fe are in a 1:1 ratio according to the balanced equation.

The molar mass of Fe2O3 can be calculated as follows:
molar mass of Fe2O3 = (atomic mass of Fe x 2) + (atomic mass of O x 3)

Looking up the atomic masses of Fe and O, we find:
atomic mass of Fe = 55.845 g/mol
atomic mass of O = 16.00 g/mol

Plugging these values into the equation:
molar mass of Fe2O3 = (55.845 g/mol x 2) + (16.00 g/mol x 3) = 159.69 g/mol

Now, we can calculate the moles of Fe2O3:
moles of Fe2O3 = 209.7 g / 159.69 g/mol = 1.311 mol

Since the ratio between Fe2O3 and CO is 1:3, we can determine the moles of CO required:
moles of CO = 3 moles of CO / 1 mole of Fe2O3 x moles of Fe2O3
moles of CO = 3 x 1.311 mol = 3.933 mol

Finally, we can calculate the mass of CO required:
mass of CO = moles of CO x molar mass of CO

The molar mass of CO can be calculated as:
molar mass of CO = atomic mass of C + atomic mass of O
atomic mass of C = 12.01 g/mol
atomic mass of O = 16.00 g/mol

Plugging these values into the equation:
molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol

Calculating the mass of CO:
mass of CO = 3.933 mol x 28.01 g/mol = 110.02 g

Therefore, approximately 110.02 grams of CO are needed to react with an excess of Fe2O3 to produce 209.7 grams of Fe.