calculate the theoretical yield (in grams) of the solid product, assuming that you use 1.0 g FeC2O4×2H2O and that oxygen is the excess reactant.

FeC2O4×2H2O (s) + O2(g) ---> FeO(s) + H2O(g) + CO2(g)

Its balanced equation is 2FeC2O4×2H2O (s) + O2(g)----> 2FeO(s) + 4H2O(g) + 4CO2(g)

Here is an example of a stoichiometry problem I've posted. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

calculate the theoretical yield (in grams) of the solid product, assuming that you use 1.0 g FeC2O4×2H2O and that oxygen is the excess reactant.

Well, I have to say, this equation definitely has a lot going on! It's like a chemistry party with FeC2O4×2H2O, O2, FeO, H2O, and CO2 all getting together for a wild ride.

Now, to calculate the theoretical yield, we need to first determine the limiting reactant. Since oxygen is in excess, we can focus on the reactant FeC2O4×2H2O.

The molar mass of FeC2O4×2H2O can be calculated as follows:

(1 mol Fe) + (2 mol C) + (4 mol O) + (4 mol H) + (2 mol O) = 215.79 g/mol

To calculate the theoretical yield, we can use the following formula:

Theoretical Yield (in grams) = (Mass of FeC2O4×2H2O / Molar mass of FeC2O4×2H2O) x (Molar ratio of FeO to FeC2O4×2H2O) x (Molar mass of FeO)

The molar ratio of FeO to FeC2O4×2H2O can be determined from the balanced equation, which is 2:2. So, it's a 1:1 ratio.

Now, let's plug in the numbers:

Theoretical Yield (in grams) = (1.0 g / 215.79 g/mol) x (1 mol FeO / 1 mol FeC2O4×2H2O) x (71.85 g/mol)

Calculating this, we get:

Theoretical Yield (in grams) = 0.014 grams of FeO

So, if everything goes according to plan, you should expect to produce approximately 0.014 grams of FeO. But remember, in chemistry, things don't always go according to plan. They can be quite unpredictable, much like a circus with clowns!

To calculate the theoretical yield of the solid product (FeO) in grams, we need to follow these steps:

Step 1: Convert the given mass of FeC2O4×2H2O to moles.
Given mass = 1.0 g
Molar mass of FeC2O4×2H2O = (1 * Molar mass of Fe) + (2 * Molar mass of C) + (4 * Molar mass of O) + (2 * Molar mass of H) + (2 * Molar mass of O)
Molar mass of Fe = 55.845 g/mol
Molar mass of C = 12.0107 g/mol
Molar mass of O = 15.999 g/mol
Molar mass of H = 1.00784 g/mol
Molar mass of FeC2O4×2H2O = (55.845) + (2 * 12.0107) + (4 * 15.999) + (2 * 1.00784) + (2 * 15.999)
Molar mass of FeC2O4×2H2O = 291.98 g/mol

Number of moles = Given mass / Molar mass
Number of moles of FeC2O4×2H2O = 1.0 g / 291.98 g/mol

Step 2: Use the balanced equation to determine the moles of FeO produced.
From the balanced equation, we know that:
2 moles of FeC2O4×2H2O = 2 moles of FeO
Number of moles of FeO = Number of moles of FeC2O4×2H2O

Step 3: Convert the moles of FeO to grams using the molar mass of FeO.
Molar mass of FeO = (1 * Molar mass of Fe) + (1 * Molar mass of O)
Molar mass of FeO = (1 * 55.845) + (1 * 15.999)
Molar mass of FeO = 71.843 g/mol

Mass = Number of moles * Molar mass
Mass of FeO = Number of moles of FeO * Molar mass of FeO

By following these steps and calculating the values, you will be able to determine the theoretical yield of the solid product (FeO) in grams.

To calculate the theoretical yield of the solid product (FeO) in grams, you need to use the given mass of the starting material (FeC2O4×2H2O) and the balanced equation.

Step 1: Convert the given mass of FeC2O4×2H2O to moles.
The molar mass of FeC2O4×2H2O can be calculated by adding the atomic masses of its constituent elements:
1 atom of Fe = 55.85 g/mol
2 atoms of C = 2 * 12.01 g/mol
4 atoms of O = 4 * 16.00 g/mol
2 formula units of H2O = 2 * (2 * 1.01 + 16.00) g/mol

Calculate the molar mass:
55.85 + 2 * 12.01 + 4 * 16.00 + 2 * (2 * 1.01 + 16.00) = 179.85 g/mol

To get moles, divide the given mass by the molar mass:
1.0 g / 179.85 g/mol = 0.00556 mol

Step 2: Use stoichiometry to convert moles of FeC2O4×2H2O to moles of FeO.
From the balanced equation, you can see that the ratio of FeC2O4×2H2O to FeO is 2:2.
So, the moles of FeO would be the same as the moles of FeC2O4×2H2O.

Therefore, the moles of FeO = 0.00556 mol.

Step 3: Convert moles of FeO to grams using the molar mass of FeO.
The molar mass of FeO can be calculated by adding the atomic masses of its constituent elements:
1 atom of Fe = 55.85 g/mol
1 atom of O = 16.00 g/mol

Calculate the molar mass:
55.85 + 16.00 = 71.85 g/mol

To get the theoretical yield in grams, multiply the moles of FeO by its molar mass:
0.00556 mol * 71.85 g/mol = 0.399 g

Therefore, the theoretical yield of the solid product (FeO) is 0.399 grams.