Combustion analysis of an 18.0 g sample of a compound yielded 9.6 g carbon, 2.0

Question

Combustion analysis of an 18.0 g sample of a compound yielded 9.6 g carbon, 2.0 g hydrogen and the remaining mass was oxygen. What would be a possible molecular formula for the compound? How many moles of A1Cl3 are produced from the reaction between 0.25 mol of Al2O3, 0.50 mol of C and 0.40 mol of Cl2 according to the reaction? Al2O3 + 3C + 3Cl2 rightarrow 3CO + 2AlCl3 Acetylsalysilic acid, aspirin, is a monoprotic acid with the chemical formula, C9H8O4 (molar mass = 180.2 g middot mol-1). What is the mass percent of acetylsalicylic acid present in a 0.1000 g tablet if 16.8 mL of 0.0200 M NaOH was required to react completely with the sample? What is the percent yield of NaN3 if 1.5 mol of NaN3 is isolated from the reaction of 6.0 mol of NaNH2 and an excess of NaN03?

Explanation / Answer

1) Let the molecular formula of the compound be CxHyOz

Now, molar mass of C = 12 g/mole

molar mass of H = 1 g/mole

molar mass of O = 16 g/mole

Now, moles of C in 9.6 g of it = 9.6/12 = 0.8

moles of H in 2 g of it = 2/1 = 2

moles of O in 6.4 g of it = 6.4/16 = 0.4

Thus, C , H & O are present in the molar ratio of 0.8:2:0.4

or, C , H & O are present in the molar ratio of 2:5:1

Thus, the empirical formula of the compound is : C2H5O

Thus, the molecular formula can be (C2H5O)n ; where n = 1,2,3.....

Hence the correct option is :- (A) C4H10O2

2) Al2O3 + 3C + 3Cl2 ----------> 3CO + 2AlCl3

Now, as per the balanced reaction:- Al2O3 , C & Cl2 reacts in the molar ratio of 1:3:3

Now, molar ratio in which Al2O3 , C & Cl2 are present = 0.25:0.5:0.4; or, 1:2:1.6 ; or, 5:10:8

Clearly, Cl2 is the limiting reagent

Thus, moles of Al2Cl3 formed = (2/3)*moles of Cl2 reacting = (2/3)*0.4 = 0.267 = 0.27

3) moles of NaOH reacting = molarity of NaOH solution*volume in litres = 0.02*0.0168 = 0.000336

Now, since aspirn is a monoprotic acid

Therefore, moles of NaOH reacting = moles of aspirin = 0.000336

Thus, mass of aspirin present = moles*molar mass of aspirin = 0.000336*180.2 = 0.06055 g

Thus, % of aspirin in the tablet = (mass of aspirin/mass of tablet)*100 = (0.06055/0.1)*100 = 60.55%

Hence the correct option is :- (C) 60.5 %

4) NaNO3 + 3NaNH2 ------------> NaN3 + 3NaOH + NH3

3 moles of NaNH2 on complete reaction produces 1 mole of NaN3

Thus, 6 moles of NaNH2 on complete reaction will produce 2 moles of NaN3

Now, moles of NaN3 actually produced = 1.5

thus, % yield = (moles of actual NaN3 produced/moles of NaN3 produced theoretically)*100 = (1.5/2)*!00 = 75%

Hence the correct option is :- (C)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---