given 40.0g of iron, how many grams of iron II oxide are produced.
51.4
To determine the number of grams of iron II oxide produced, we need to understand the chemical equation for the reaction. Assuming that iron reacts with oxygen to form iron II oxide, the balanced equation is:
4 Fe + 3 O2 -> 2 Fe2O3
From the balanced equation, we see that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron II oxide. To find the mass, we need to convert the given mass of iron into moles using the molar mass of iron (Fe), which is 55.85 g/mol.
Step 1: Calculate the moles of iron (Fe) given:
moles of Fe = mass of Fe / molar mass of Fe = 40.0 g / 55.85 g/mol ≈ 0.716 mol
Step 2: Determine the moles of iron II oxide (Fe2O3) produced based on the balanced equation:
moles of Fe2O3 = (moles of Fe / 4) * 2 = (0.716 mol / 4) * 2 ≈ 0.358 mol
Step 3: Convert the moles of iron II oxide to grams using the molar mass of Fe2O3, which is equal to 159.69 g/mol (56.00 g/mol for iron * 2 + 16.00 g/mol for oxygen * 3):
grams of Fe2O3 = moles of Fe2O3 * molar mass of Fe2O3 = 0.358 mol * 159.69 g/mol ≈ 57.15 g
Therefore, approximately 57.15 grams of iron II oxide are produced from 40.0 grams of iron.
balance the equation:
2Fe + O2 >> 2FeO
so for each mole of Fe, you get the same number of moles of FeO
moles Fe= 40/atomicmassFe