Laboratory Manual for General Organic, and Biochemistry, Seventh Edition 127 Nam

Question

Laboratory Manual for General Organic, and Biochemistry, Seventh Edition 127 Name Locker Number Date Phease print lest name irat PRELIMINARY EXERCISES: Experiment 9 Analysis of a KCio,-KCI Mixture 1. Aluminum powder can burn in pure oxygen to form aluminum oxide, Alo The balanced equation is: Write numbers in the small blanks to identify four mole ratios that exist in this balanced equation. You may want to refer to the Chemical Arithmetic-Equations exercises: mol Al mol Al - mol O, Referring to the balanced equation in question 1, what mass of aluminum is required to mol Al,0, mol Al mol O2 mol Al,o- mol Al,o, 2. just completely react with 96.0 g of O, The molar masses are: Al-27.0 &O;, -32.0 g Please show your calculations below Mass of Al = How many grams of oxygen, O, would be produced in the complete decomposition of 10.0 g of KCIo,? The molar masses are: o, -32.0 & KCIO, equation is: 3. -122.6& The balanced MnO, Mass of O,

Explanation / Answer

4Al(s) + 3O2(g) ----------> 2Al2O3(s)
4 moles of Al/3 moles of O2   2 moles of Al2O3/4 moles of Al 3 moles of O2/2 moles of Al2O3
4 moles of AL/2 moles of Al2O3
2. 4Al(s) + 3O2(g) ----------> 2Al2O3(s)
from balanced equation
   3 moles of O2 react with 4 moles of Al
   3*32g of O2 react with 4*27g of Al
   96g of O2 react with = 4*27*96/3*32 = 108 g of Al >>>>>answer
3. 2KClO3 ------> 2KCl + 3O2
2 moles of KClO3 decomposes to gives 3 moles of O2
2*122.6g of KClO3 decomposes to gives 3*32g of O2
   10g of KClO3 decomposes to gives = 3*32*10/2*122.6   = 3.915g of O2

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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