C × () MyLab ind Mastering Course Home × i? courseld 1 4 1 702298OpenVellumh MAC

Question

C × () MyLab ind Mastering Course Home × i? courseld 1 4 1 702298OpenVellumh MAC. 3356dS7essdessese48b1c4e3ce63. 10001 hemist Ch. Fons Hw Secure https/ ngchemistry.com/myct/itemView?assignmentProblemiDs 91174904 Chemis Question 18 urses previous 18 of 21 ned e Home Question 18 Part A bus ron1l) oxide reacts with carbon monaxide according to the equation Once the reaction has occurred as completely as possible what mass (in g) of the excess neactant is left? A reaction mixture initially contains 22 05 g FeO and 14.22 g CO res er Settings My Answers Give p aeea

Explanation / Answer

A)

Molar mass of Fe2O3,

MM = 2*MM(Fe) + 3*MM(O)

= 2*55.85 + 3*16.0

= 159.7 g/mol

mass(Fe2O3)= 22.05 g

number of mol of Fe2O3,

n = mass of Fe2O3/molar mass of Fe2O3

=(22.05 g)/(159.7 g/mol)

= 0.1381 mol

Molar mass of CO,

MM = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass(CO)= 14.22 g

number of mol of CO,

n = mass of CO/molar mass of CO

=(14.22 g)/(28.01 g/mol)

= 0.5077 mol

Balanced chemical equation is:

Fe2O3 + 3 CO ---> 2 Fe + 3 CO2

1 mol of Fe2O3 reacts with 3 mol of CO

for 0.1381 mol of Fe2O3, 0.4142 mol of CO is required

But we have 0.5077 mol of CO

so, Fe2O3 is limiting reagent

we will use Fe2O3 in further calculation

According to balanced equation

mol of CO reacted = (3/1)* moles of Fe2O3

= (3/1)*0.1381

= 0.4142 mol

mol of CO remaining = mol initially present - mol reacted

mol of CO remaining = 0.5077 - 0.4142

mol of CO remaining = 0.0935 mol

Molar mass of CO,

MM = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass of CO,

m = number of mol * molar mass

= 9.346*10^-2 mol * 28.01 g/mol

= 2.62 g

Answer: 2.62 g

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.