Question

For the following reaction, 0.103 moles of iron are mixed with 0.302 moles of oxygen gas. iron(s) + oxygen (g) → iron(II) oxide(s) What is the formula for the limiting reagent? What is the maximum amount of iron(II) oxide that can be produ moles

Answer 1

Iron (8) + Oxygen (a) → Iron (0) oxide (s) a Fe(s) + O2 (8) Fe Oz (5) from the above balanced reaction equation, it is clear that I mole of Fe (s) reacts with a mole of O₂ to give a m clear that, uation, it is fel 0.103 " 00103 " un " 0.103" 0.103 - than There are 8.302 0.302 mole mole of of oxygen which oxygen which is greater the amount required to completed reaction with Fe(s). So, here, the formula of limiting reagent s B iron(s) or Fe(s). Aug amount of iron(a) oxide that - maximum can be is - 0.403 more. Ang produced - The maximum an anced re 9 сня ( + 2с14 ( 5 2CH, C, Сә) [kalar . equation) from the above balanced reaction equations, & more of Chalos reacts with mole CH22) a mole of cela (s) to give a 0.275 " " 0.275 m an n o 2x0.245 " "n But there are 0.304 moles of cela los which is in excess. : When Oiaas mole of methane is mixed with 0.309 more of carbon tetrachloride, the formula for limiting reagent is - CHa Cos. Ang : The maximum amount of lichloromethane (Chade can be produced = 2x0.275 mole = 0.550 mole. Ang