5. How many grams of iron oxide can be produced from 2.50 g of oxygen reacting with iron, according to the following equation?

2 Fe (s) + 3 O2 (g) -->2 Fe2O3(s)

the molar mass of O₂ is 32 g

3 moles of oxygen produce 2 moles of iron oxide

(2.50 / 32)* 2/3 = moles of Fe₂O₃

the molar mass of Fe₂O₃ is 159.7 g

88 g

To find out how many grams of iron oxide can be produced from 2.50 g of oxygen reacting with iron, we will use stoichiometry and the balanced chemical equation.

First, let's calculate the molar mass of Fe2O3 (iron oxide). The molar mass of iron (Fe) is 55.85 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol. Since there are two iron atoms in Fe2O3 and three oxygen atoms, we can calculate the molar mass as follows:

Fe2O3 = (2 * 55.85 g/mol) + (3 * 16.00 g/mol)
= 111.70 g/mol + 48.00 g/mol
= 159.70 g/mol

According to the balanced chemical equation, 2 moles of Fe2O3 are produced for every 3 moles of O2. This means that the ratio of the moles of Fe2O3 to O2 is 2:3.

Now, let's calculate the number of moles of oxygen (O2). The molar mass of O2 is 32.00 g/mol. We can use the following equation to calculate the moles:

moles = mass / molar mass

moles of O2 = 2.50 g / 32.00 g/mol
= 0.0781 mol

Using the ratio of 2:3, we can determine the moles of Fe2O3 produced:

moles of Fe2O3 = (2/3) * 0.0781 mol
= 0.0521 mol

Finally, we can calculate the mass of Fe2O3 produced using its molar mass:

mass of Fe2O3 = moles of Fe2O3 * molar mass
= 0.0521 mol * 159.70 g/mol
= 8.31 g

Therefore, 2.50 g of oxygen reacting with iron will produce approximately 8.31 g of iron oxide.