{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework #1 Solutions

# Homework #1 Solutions - Chemistry 30.101 Homework#1 —...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chemistry 30.101 Homework #1 — Solution l. 11B at # = 5 ——> 5 protons and 5‘electrons; 11— 5 = 6 neutrons 12C at # = 6 —> 6 protons and 6 electrons; 12 — 6 = 6 neutrons 13C at # = 6 —> 6 protons and 6 electrons; 13 - 6 = 7 neutrons 20Ne at # = 10 ——> 10 protons and 10 electrons; 20 - 10 = 10 neutrons 2831 at # = 14 —> 14 protons and 14 electrons; 28 - 14 = 14 neutrons 2. C(4 beyond He) 2 4, Na(1 beyond Ne) = 1, Si(4 beyond Ne) : 4, N(5 beyond He) = 5 P(5 beyond Ne) = 5, K(1 beyond Ar) = 1 (Na, K) both with 1, (C, Si) both with 4, (N, P) both with 5 3. mW(H202) = 2x1+ 2x16 = 2 + 32 = 34 mw(C0) = 12 + 16 = 28 mw(C5H6) = 6x12 + 6x1 = 72 + 6 = 78 mw(C2H5OH) = 2x12 + 6x1 + 16 = 24 + 6 + 16 = 46 4. 1 mole of gold weighs 196.97 grams (at Wt of Au) (grams/mole)/(atoms/mole) = 196.97/6.022x1023 (grams/atom) = 32,71x10'23 = 3.2711(10‘22 1 mole of C02 weights 44 grams (mw of CO2) (grams/mole)/(moleculcs/mole) = 44/6.022x1023 (grams/molecule) = 7.31x10‘23grams/molecule 5. moles = grams/mw moles(H202) = 10/34 = 0.294 moles(CO) = 10/28 = 0.357 moles(C6H6) : 10/78 = 0.128 moles(C2H50H) = 10/46 = 0.217 6. grams 2 moles X mw , grams(H202) = (1/4) (34) = 8.5 grams(C0) = (1/4) (28) = 7 grams(C6H5) = (1/4) (78) = 19.5 grams(C2H50H) = (1/4) (46) = 11.5 7. C2H6 + (7/2) 02 —-> ZCOZ + 3H20 (2C,6H,70) (2C,6H,70) 8. CH3OH + (3/2) 02 -—> C02 + 21—120 (1C,4H,40) (1C,4H,40) 9. mw C2H6 = (2x12 + 6x1) = 24 + 6 = 30, moles C2H6 : grams/mw = 15/30 = 1/2 From the stoichiometry in (7) 1 mole C2H5 ——> 2 moles C02 and 3 moles H20 so (1/2) mole C2H6 —> 1 mole C02 and (3/2) moles H2O mw C02 2 (12 + 2x16) = 44 so 1 mole C02 —-> 44 grams C02 mw H20 = (2x1 + 16) = 18 so 3/2 moles H20 —> 27 grams H20 10. mw CH30H = (12 + 4x1 + 16) = 32, mw 02 : (2x16) = 32 [same mw] 16 grams of each —> 1/2 mole of each So one has the same number of moles of each, but from the stoichiometry in (8) the reaction requires (3/2) as many molecules of 02 as of methanol. Thus one does not have enough 02 to react with all of the methanol and hence 02 is the limiting reagent in this case. One can rewrite (8): (2/3) CH30H + 02 —> (2/3) C02 + (4/3)H20 Then 1/2 moles of 02 —> (1/2)(2/3) = (1/3) moles of C02 4 (44/3) = 14.7 grams of C02 1/2 moles of 02 -—> (1/2)(4/3) = (2/3) moles of H20 —9 (2/3)(18) = 12 grams H20 One has (2/3)(1/2) = 1/3 moles of CH30H used up, so there are (1/2) — (1/3) = 1/6 moles of methanol left unreacted. ...
View Full Document

{[ snackBarMessage ]}