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.

All the elements are arranged by melting points on the periodic table. in row they go from metals to gasses. in the row from sodium to argon, sodium is more reactive than the one on its right, magnesium.