Michael S. answered • 13d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
Short answers first. (a) 5.66 g of C, (b) 90.9 g of CO, (c) 4.34 g of Fe.
All three parts ride the same road: grams -> moles -> mole ratio from the balanced equation -> grams. The equation is already balanced for you, so every coefficient you need is sitting right there.
Fe2O3(s) + 3 C(s) -> 2 Fe(s) + 3 CO(g)
Molar masses (use whatever your own table gives; these are five figures):
Fe2O3 = 2(55.845) + 3(15.999) = 159.69 g/mol C = 12.011 g/mol CO = 12.011 + 15.999 = 28.010 g/mol Fe = 55.845 g/mol
Part (a) - grams of C needed to react with 25.1 g of Fe2O3
25.1 g Fe2O3 / 159.69 g/mol = 0.15718 mol Fe2O3 0.15718 mol Fe2O3 x (3 mol C / 1 mol Fe2O3) = 0.47154 mol C 0.47154 mol C x 12.011 g/mol = 5.66 g C
Part (b) - grams of CO produced when 39.0 g of C reacts
39.0 g C / 12.011 g/mol = 3.2470 mol C 3.2470 mol C x (3 mol CO / 3 mol C) = 3.2470 mol CO 3.2470 mol CO x 28.010 g/mol = 90.9 g CO
Two things about part (b) are worth stopping on. First, that ratio is 3 to 3, which is 1 to 1 - the moles pass straight through untouched. Seeing two 3s in the equation and multiplying by 3 anyway is the most common error on this part, and it gives 273 g. Write the ratio as a fraction with the units attached and it cancels itself correctly.
Second, 39.0 g of carbon turns into 90.9 g of carbon monoxide - the mass more than doubles, and that is not a mistake. Every carbon atom picks up an oxygen on its way to CO, and those oxygens come out of the Fe2O3. Mass is conserved across the whole reaction, never across one reactant on its own. If you expected the product mass to be smaller than 39.0 g, that instinct is the thing to fix here.
One honest caveat on this part: it lands right on a rounding boundary. With 12.011 and 28.010 you get 90.949, which reports as 90.9. If your book rounds carbon to 12.01 you get 90.96, which reports as 91.0. Both are defensible - use the masses your course gives you, and do not be rattled if the key says 91.0.
Part (c) - grams of Fe produced from 6.20 g of Fe2O3
6.20 g Fe2O3 / 159.69 g/mol = 0.038825 mol Fe2O3 0.038825 mol x (2 mol Fe / 1 mol Fe2O3) = 0.077650 mol Fe 0.077650 mol Fe x 55.845 g/mol = 4.34 g Fe
The free cross-check nobody teaches: conservation of mass. Take part (a). You put in 25.1 g of Fe2O3 plus the 5.66 g of C it demands, so 30.76 g goes in. Now build both products off that same 0.15718 mol:
Fe: 0.15718 x 2 x 55.845 = 17.56 g CO: 0.15718 x 3 x 28.010 = 13.21 g total out = 30.77 g
That reproduces the 30.76 g that went in, to rounding. It costs one line and it catches a wrong molar mass, an upside-down ratio or a dropped coefficient on the spot. Run it any time a problem gives you enough to close the loop.
A second, faster check on part (c): it starts from the same reactant as part (a), so the two answers have to be proportional. 6.20 / 25.1 = 0.2470, and 0.2470 x 17.56 g = 4.34 g. Same answer, no new arithmetic.
Three traps to watch. (1) The three parts are independent. Each one hands you fresh data, so nothing carries forward from (a) into (b) or (c). Dragging an earlier answer along is a clean way to lose all three at once. (2) Fe2O3 is 159.69 g/mol, not 111.69. Forgetting the three oxygens quietly wrecks (a) and (c) together, and it always makes your answer come out too big. (3) The mole ratio is want-over-have. Write it as a fraction, units included, and let them cancel - if "mol Fe2O3" does not cancel out, the fraction is upside down.
Last thing, on significant figures: 25.1, 39.0 and 6.20 each carry three, so all three answers report to three. Carry the extra digits through the intermediate steps and round only once, at the end. Rounding 0.15718 to 0.157 partway through will start nudging your last digit around, and part (b) above shows how little it takes to cross a boundary.
