Question

How many grams of C are required to react with 43.8 g of Fez O3?

Answer 1

In order to determine the number of grams of C required to react with 43.8 g of Fe2O3, we need to write and balance the equation for the reaction.

The balanced equation for the reaction between C (carbon) and Fe2O3 (iron(III) oxide) is:
C + Fe2O3 -> CO2 + 2Fe

From the balanced equation, we can see that 1 mole of C reacts with 1 mole of Fe2O3 to produce 1 mole of CO2 and 2 moles of Fe.

Step 1: Calculate the molar mass of Fe2O3.
Fe2O3 = (2 * atomic mass of Fe) + (3 * atomic mass of O)
Fe2O3 = (2 * 55.845 g/mol) + (3 * 16.00 g/mol)
Fe2O3 = 111.69 g/mol + 48.00 g/mol
Fe2O3 = 159.69 g/mol

Step 2: Convert the mass of Fe2O3 to moles.
moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
moles of Fe2O3 = 43.8 g / 159.69 g/mol
moles of Fe2O3 = 0.274 moles

Step 3: Determine the number of moles of C required.
From the balanced equation, we know that 1 mole of C reacts with 1 mole of Fe2O3. Therefore, the number of moles of C required is also 0.274 moles.

Step 4: Convert moles of C to grams.
mass of C = moles of C * molar mass of C
mass of C = 0.274 moles * 12.01 g/mol
mass of C = 3.29 grams (rounded to two decimal places)

Therefore, approximately 3.29 grams of C are required to react with 43.8 g of Fe2O3.