How many grams of iron is formed by the interaction of 4.64 g of iron scale (Fe3O4) and 2.24 liters of carbon monoxide (carbon monoxide (2))
Let’s execute the solution:
1. Let’s compose the equation:
Fe3O4 + 4CO = 3Fe + 4CO2 – ОВР, iron obtained;
2. Let’s make calculations using the formulas:
M (Fe3O4) = 231.4 g / mol;
M (CO) = 28 g / mol;
M (Fe) = 55.8 g / mol.
3. Determine the amount of moles of iron scale, carbon monoxide (2):
Y (Fe3O4) = m / M = 4.64 / 231.4 = 0.02 mol (deficient substance);
1 mol of gas at normal level – 22.4 liters.
X mol (CO) – – 2.24 l. hence, X mol (CO) = 1 * 2.24 / 22.4 = 0.1 mol (substance in excess);
Calculations are made taking into account the deficient substance.
4. Let’s make the proportion:
0.02 mol (Fe3O4) – X mol (Fe);
-1 mol -3 mol from here, X mol (Fe) = 0.02 * 3/1 = 0.06 mol.
5. Find the mass of iron:
m (Fe) = Y * M = 0.06 * 55.8 = 3.348 g. = 3.35 g.
Answer: the mass of iron is 3.35 g.
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