Question

kec Use the Relerences to access important valees if areded for this question For the following reaction, 25.0 grams of iron are allowed to react with 6.19 grams of exygen gas iron (s) + oxygen (ro(I oxide (s) What is the maximum amount of iron(II oxide that can be formed? What is the FORMULA for the limiting reagent? grams What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 7 more group attempts remaining

Answer 1

The balanced equation is

2 Fe + O₂ -----> 2 FeO

Number of moles of Fe = 25.0 g / 55.845 g/mol = 0.448 mol

Number of moles of O2 = 6.19 g / 32.0 g/mol = 0.193 mol

From the balanced equation we can say that

2 mole of Fe requires 1 mole of O2 so

0.448 mole of Fe will require

= 0.448 mole of Fe *(1 mole of O2 / 2 mole of Fe)

= 0.224 mole of O2

But we have 0.193 mole of O2 which is in short so O2 is limiting reactant

The formula of limiting reactant is O₂

From the balanced equation we can say that

1 mole of O2 produces 2 mole of FeO so

0.193 mole of O2 will produce

= 0.193 mole of O2 *(2 mole of FeO / 1 mole of O2)

= 0.386 mole of FeO

mass of 1 mole of FeO = 71.844 g

so the mass of 0.386 mole of FeO = 27.7 g

Therefore, the mass of FeO produced would be 27.7 g

The number of moles of excess reactant = 0.448 - 0.386 = 0.0620 mole

mass of 1 mole of Fe = 55.845 g so

the mass of 0.0620 mole of Fe = 3.46 g

Therefore, the mass of excess reactant remain = 3.46 g