If 3.18 x 1023 atoms of iron react with 67.2 L of chlorine gas at STP, what is the maximum number of grams of iron (III) chloride that can be produced?

so, you have

3.18/6.02 = 0.528 moles of Fe
67.2/22.4 = 3 moles of Cl2
Now look at your equation to see which is the limiting reagent, and thus determine the moles and grams of product.

To find the maximum number of grams of iron (III) chloride that can be produced, we need to use stoichiometry and convert the given quantities to moles, and then use the balanced chemical equation to determine the molar ratio between iron (III) chloride and iron.

First, let's write the balanced chemical equation for the reaction between iron and chlorine gas to form iron (III) chloride:

2Fe + 3Cl2 -> 2FeCl3

According to the balanced chemical equation, 2 moles of iron react with 3 moles of chlorine gas to produce 2 moles of iron (III) chloride.

Since we are given the number of atoms of iron (3.18 x 10^23 atoms), we need to convert it to moles. We know that 1 mole of any substance contains 6.022 x 10^23 particles (Avogadro's number).

So, the number of moles of iron is:

moles of iron = (3.18 x 10^23 atoms) / (6.022 x 10^23 atoms/mol) = 0.528 mol

Next, we need to convert the volume of chlorine gas (67.2 L) to moles using the ideal gas law at STP (Standard Temperature and Pressure).

At STP, 1 mole of any ideal gas occupies 22.4 L of volume.

So, the number of moles of chlorine gas is:

moles of chlorine gas = (67.2 L) / (22.4 L/mol) = 3 mol

Now, we can determine the limiting reagent. Since the ratio between iron and chlorine gas in the balanced equation is 2:3, it means that for every 2 moles of iron, we need 3 moles of chlorine gas. The limiting reagent is the reactant that is completely consumed and limits the amount of product formed.

In this case, since we have 0.528 mol of iron and 3 mol of chlorine gas, iron is in excess, and chlorine gas is the limiting reagent.

Now, we need to calculate the number of moles of iron (III) chloride produced using the stoichiometry of the balanced equation.

Since the molar ratio between iron (III) chloride and chlorine gas is 2:3, it means that for every 3 moles of chlorine gas reacted, we produce 2 moles of iron (III) chloride.

moles of iron (III) chloride = (3 mol of chlorine gas) * (2 mol of FeCl3 / 3 mol of Cl2) = 2 mol

Finally, we can calculate the mass of iron (III) chloride produced using the molar mass of FeCl3.

The molar mass of FeCl3 is:

molar mass of FeCl3 = (1 mole of Fe x atomic mass of Fe) + (3 moles of Cl x atomic mass of Cl)

Using the atomic masses of iron (Fe = 55.85 g/mol) and chlorine (Cl = 35.45 g/mol), we can calculate the molar mass of FeCl3:

molar mass of FeCl3 = (1 mol x 55.85 g/mol) + (3 mol x 35.45 g/mol) = 162.2 g/mol

Finally, we can calculate the maximum mass of iron (III) chloride produced:

mass of FeCl3 = (2 mol of FeCl3) * (162.2 g/mol) = 324.4 g

Therefore, the maximum number of grams of iron (III) chloride that can be produced is 324.4 grams.