The reaction between potassium superoxide, KO_2, and CO_2, 4KO_2 + 2CO_2 right a

Question

The reaction between potassium superoxide, KO_2, and CO_2, 4KO_2 + 2CO_2 right arrow 2K_2CO_3 + 3O_2 is used as a source of O_2 and absorber of CO_2 in self-contained breathing equipment used by resque workers. (Figure 1) How many moles of O_2 are produced when 0.475 mol of KO_2 reacts in this fashion? Express your answer with the appropriate units. N_(O_2) = How many grams of KO_2 are needed to from 7.0 g of O_2? Express your answer with the appropriate units. M(KO_2) = How many grams of CO_2 are used when 7.0 g of O_2 are produced? Express your answer with the appropriate units. M(CO_2) =

Explanation / Answer

KO2 = 0.475 mol

ratio is 4:3 so

0.475*3/4 = 0.35625 mol of O2

b.

mol of O2 = mass/MW = 7/32 = 0.21875 mol of O2

ratio is 4:3

so

4/3*0.21875 = 0.29166 mol of KO2

mass = mol*MW = 71.10*0.29166 = 20.737026 g of KO2

c.

ratio is 3:2 so

2/3*0.29166 = 0.1944mol of CO2

mnass = mol*MW = 0.1944*44 = 8.5536 g of CO2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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