An iron nail rusts when exposed to oxygen. How many grams of iron are needed to completely consume 26.7 L of oxygen gas according to the following reaction at 25 °C and 1 atm?

Answer 1

First we write out the balanced equation. We assume that the iron nail is completely made of pure iron (Fe). We also need to know the chemical compound that corresponds to iron rust; the chemical compound is Fe₂O₃ (which is also known as iron (III) oxide). Since the problem says that the nail rusts "when exposed to iron", we know that the iron (in the nail) and the O₂ are the reactants, and the rust (Fe₂O₃) is the product. So the unbalanced chemical equation looks like this:

Fe (s) + O₂ (g) --> Fe₂O₃ (s)

We must balance this equation; the balanced chemical equation is:

4 Fe (s) + 3 O₂ (g) --> 2 Fe₂O₃ (s)

The problem says that we have 26.7 L of oxygen gas. First we need to figure out how many moles of oxygen gas are in 26.7 L of oxygen gas. To figure this out, we must use the ideal gas equation, which is:

PV = nRT

In the ideal gas equation, P stands for pressure. The problem told us that the pressure is 1 atm. So P = 1 atm. The V stands for volume, and the problem told us we have 26.7 L of oxygen gas, so V = 26.7 L of O2. The n stands for moles of O2, which is what we want to find. The R stands for the ideal gas constant, which you can look up in a chart or table. For this problem, we will use R = 0.082057 L*atm/(K*mole). Finally, T stands for temperature in Kelvin. The problem said the temperature was 25 degrees Celsius. We have to convert the temperature to Kelvin. We convert using the equation K = 273.15 + C. So, K = 273.15 + 25 = 298.15. So the temperature = 298.15 K. Now we put these values into the ideal gas equation, and it looks like this:

1 * 26.7 = n * 0.082057 * 298.15

First, we can simplify 1 * 26.7 to just 26.7. So the equation looks like:

26.7 = n * 0.082057 * 298.15

Now we multiply 0.082057 * 298.15. The calculator says that 0.082057 * 298.15 = 24.4652946. So our equation now looks like this:

26.7 = n * 24.4652946

Finally, we just divide both sides by 24.4652946:

n = 26.7 / 24.4652946

The calculator tells us that 26.7 / 24.4651946 = 1.09134632. So, there are 1.09134632 moles of O₂. Now we convert this 1.09134632 moles of O2 to moles of iron. We do this using the coefficients in the balanced equation. The balanced equation is: 4 Fe (s) + 3 O₂ (g) --> 2 Fe₂O₃ (s) . The coefficient of O2 in the balanced equation is 3. The coefficient of Fe in the balanced equation is 4. The equation to convert from moles of O2 to moles of Fe looks like this:

moles of O2 x coefficient of Fe

coefficient of O2

In case you can't see this, it is the moles of O2 multiplied by (the coefficient of Fe / the coefficient of O2). Remember we just found the moles of O2 (1.09134632 moles of O2), and we just stated the coefficients we needed. So with these values, the equation to convert from moles of O2 to moles of Fe looks like this:

1.09134632 x 4

3

In case you can't see this, we do 1.09134632 x (4/3).

The calculator says that 1.09134632 x (4/3) = 1.45512842. So there are 1.45512842 moles of Fe. But the problem wants grams of Fe. So we need to turn 1.45512842 moles of Fe into grams of Fe. We do this by finding the molar mass of Fe. We look on the periodic table and see that the atomic mass of Fe is 55.847. So 1 mole of Fe = 55.847 grams of Fe. To convert the 1.45512842 moles of Fe to grams of Fe, we do 1.45512842 x 55.847. The calculator says that 1.45512842 x 55.847 = 81.264557. So, there are 81.264557 grams of Fe in 1.45512842 moles of Fe. We should round our answer to 3 significant figures since the number that the problem gave us (26.7) has 3 significant figures. If we round 81.264557 to 3 significant figures, we get 81.3. So the final answer is 81.3 grams of Fe.