How many grams of co are needed with an excess of fe2o3 to produce 27.9 g fe? show work. fe2o3(s)+3co(> 2fe(s)+3co2(g)

21.0 g.

Explanation:

Firstly, It is considered as a stichiometry problem.From the balanced equation: Fe₂O₃ + 3CO → 2Fe + 3CO₂.It is clear that the stichiometry shows that 1.0 moles of Fe₂O3 reacts with 3.0 moles of CO to give 2.0 moles of Fe and 3.0 moles of CO₂.We must convert the grams of Fe (27.9 g) to moles (n = mass/Atomic mass).n = (27.9 g) / (55.845 g/mole) = 0.499 ≅ 0.50 moles.Now, we can get the number of moles of CO that is needed to produce 0.50 moles of Fe.Using cross multiplication:3.0 moles of CO → 2.0 moles of Fe, from the stichiometry of the balanced equation.??? moles of CO → 0.50 moles of FeThe number of moles of CO that will produce 0.50 moles of Fe (27.9 g) is (3.0 x 0.50 / 2.0) = 0.75 moles.Finally, we should convert the number of moles of CO into grams (n = mass/molar mass).Molar mass of CO = 28.01 g/mole.mass = n x molar mass = (0.75 x 28.01) = 21.0 g.

Well technically there are two correct answers b and d. but for you its probably b in general.good day sir/ma'am : ) (i know not gender)

Hello! data: p (pressure) = 1.25 atmv (volume) = ? (in l)t (temperature) = 298 kn (number of mols) = 0.900 molesr (gas constant) = 0.082 (atm*l/mol*k)apply the data to the clapeyron equation (ideal gas equation), see: i hope this . =)