Read each question and grading rubric thoroughly. Be sure to include units on nu

Question

Read each question and grading rubric thoroughly. Be sure to include units on numerical answers, to the correct number of significan t figures, where appropriate. 1. Consider the unbalanced equation below: L,os() FCo(8) )+) unbalanced (a) (2 pts) What element is being oxidized? OI OC oo O nothing (b) Balance the equation above using the half- reaction method in acidic conditions. Include state symbols! i. (6 pts) Balanced Oxidation half-reaction ii. (6 pts) Balanced Reduction half-reaction ili. (6 pts) Balanced overall equation for this reaction, simplified, based on your answers above. Show work Simplified: (c) (10 pts) If 10.53 g of 120s was allowed to react with 0.575 g of CO, how many grams of 120s would be leftover?

Explanation / Answer

Ans 1 :

a) C

Carbon is getting oxidised. Its oxidation state has been increased from +2 to +4.

b)

i) Balanced oxidation half reaction :

5CO + 5H2O = 5CO2 + 10H+ + 10e-

ii) Balanced reduction half reaction :

I2O5 + 10H+ + 10e- = I2 + 5H2O

iii) The balanced overall reaction is given as :

5CO + I2O5 + 5H2O + 10H+ + 10e- = 5CO2 + I2 + 5H2O + 10H+ + 10e-

Simplified :

5CO + I2O5 = I2 + 5CO2

c)

10.53 g of I2O5 = 10.53 / 333.8 = 0.032 moles

0.575 g of CO = 0.575 / 28.01 = 0.020 moles

0.032 moles of I2O5 will require 0.032 x 5 = 0.16 moles of CO , so CO is the limiting reagent

0.020 moles of CO will require 0.020 / 5 = 0.004 moles of I2O5

Mass of I2O5 utilised = 0.004 x 333.8 = 1.37 grams

So mass of I2O5 left unreacted = 10.53 - 1.37

= 9.16 grams

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

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