What is the balanced equation for the combustion of 1-naphthol, C_10 H_7 OH, to

Question

What is the balanced equation for the combustion of 1-naphthol, C_10 H_7 OH, to give CO_2 and H_2 O? Do not include phases. How many miligrams of CO_2 and H_2 O are produced from the complete combustion of 5.135 mg of C_10 H_7 OH? A 10.05 g sample of Mo_2 O_3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 10.722 g. Add subscripts below to correctly identify the empirical formula of the new oxide. Calculate the number of grams of xenon in 4.465 g of the compound xenon tetrafluoride. Prior to their phaseout in the 1980s, chemicals containing lead were commonly added to gasoline as anti-knocking agents. A 4 485 g sample of one such additive containing only lead, carbon, and hydrogen was burned in an oxygen rich environment. The products of the combustion were 4.882 g of CO_2 (g) and 2.498 g of H_2 O(g). Insert subscripts below to appropriately display the empirical formula of the gasoline additive:

Explanation / Answer

1) The balanced equation for combustion of 1-naphthol is

C10H7OH + 23/2 O2  -------->10 CO2 + 4H2O

molar mass of naphthol is 144g/mol

Thus one mol of naphthol gives 10 moles of CO2

144 g of naphthol gives 10x 44 g of CO2

or 144 mg of naphthol gives 10 x 44 mg of CO2

5.135 mg of naphthol gives 15.69 mg of CO2

Similarly

144 g of naphthol gives 4x18g of H2O

or 144m g of naphthol gives 4x18mg of H2O

thus  5.135 mg of naphthol gives 2.567mg of H2O

Q2)

Mo2O3 is oxidised to another oxde of molybdenum.

Thus we can say equivalents of initial oxide = equivalents of other oxide formed

equivalents of Mo2O3 = weight/ eq. wt of it = 10.05g / (95.94/3) +8 =0.2513

atomic mass of Mo is 95.94g/mol and equivalent weight of oxygen is 16/2 =8g/eq

similarly equivalents of oxide formed = 10.722g/ (96.94/x +8)

by equating and solving for x we get x= 2.767

Thus empirical formula is Mo2O2.767 or Mo4O11

Q3) Atomic mass of xe is 131.29 g/mol and F is 19

Thus molar mass of XeF4bis 207.3 g/mol of which xenon has 131.3 g

Hence 4.465g of compound can have xenon of 2.828g

Q4)

% oc carbon = (wt. of CO2 x12 x100) / ( wt of compoundx44)

= 4.882x12x100 ]/ (4.485x44) = 29.686

% of H = [wt. of H2O x2 x100]x[wt of compoundx18]

= 6.1885

Thus % of lead = 64.125

element C H Pb

% 29.686 6.188 64.125

At.Wt. 12 1 206

number of moles 2.47 6.188 0.311

ratio ofmoles 7.95 19.95 1

Thus the empirical formula is C8H20Pb or Pb (C2H5)4

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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