Question

For this reaction, 26.0 g iron reacts with 6.11 g oxygen gas. iron (s) + oxygen (g) iron(II) oxide (s) What is the maximum mass of iron(II) oxide that can be formed? g What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete?

Answer 1

Balanced equation for the reaction of iron and oxygen gas

2Fe(s) + O2(g) ------> 2FeO(s)

Mass of Fe = 26.0 g, Atomic weight of Fe = 55.845 g/mol

Number of moles = Mass in gram / Atomic weight

Number of moles of Fe = 26.0 g / 55.845 g/mol = 0.4656 mol

Mass of O2 = 6.11 g, Molar mass of O2 = 32 g/mol

Number of moles of O2(g) = 6.11 g / 32 g/mol = 0.1909 mol

From reaction, 1.0 mol of O2 reacts with 2.0 mol of Fe produces 2.0 mol of FeO then 0.1909 mol of O2 will react with 2.0 * 0.1909 mol = 0.3818 mol of Fe will produce 0.3818 mol of FeO.

Molar mass of FeO = 71.844 g/mol

Maximum mass of iron(II)oxide can be produced = 71.844 g/mol * 0.3818 mol = 27.43 g

Since O2(g) consumes compltely during the reaction hence limiting reagent formula is O2.

Excess moles of Fe remaining = (0.4656 - 0.3818) mol = 0.0838 mol

Mass of Fe reagent remains after the reaction = 0.0838 mol * 55.845 g/mol = 4.68 g