prelim1ans_chem209_fall2007

# Prelim1ans_chem209_f - CHEMISTRY 209 PRELIM I October 4 2007 ANSWER KEY 1 1(a(7 points Balance the following chemical reaction B5 H9(g O2(g B2 O3(s

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CHEMISTRY 209 PRELIM I October 4, 2007 ANSWER KEY 1

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1. (a) (7 points) Balance the following chemical reaction: B 5 H 9 (g) + O 2 (g) −→ B 2 O 3 (s) + H 2 O(g) First, work on the B; there are 5/formula unit on the left, 2/formula unit on the right; so we need 2 moles on the left and 5 on the right. That gives us 18 H units on the left, that we need to account for on the right; call it 9 H 2 O units. In total, that means we have 15 + 9 = 24 oxygen atoms on the right that we need to account for on the left; or 12 O 2 units. Ending up with: 2B 5 H 9 ( g )+ 12O 2 ( g ) 5B 2 O 3 ( s 9H 2 O ( g ) (b) (8 points) The hormone adrenaline has the molecular formula C 9 H 13 O 3 N. Complete combustion of adrenaline yields only CO 2 ,H 2 OandN 2 as products. Balance the chemical equation below representing the combustion of adrenaline by Flling in the blanks with appropriate coeﬃcients. C 9 H 13 O 3 N+ O 2 CO 2 + H 2 O+ N 2 Balance the equation by Fnding the coeﬃcients which belong in the blanks. Approach as in the previous problem. Each C on the left has to end up in a CO 2 molecule on the right; each 2 H’s on the left in an H 2 O molecule on the right, and each N atom as a part of N 2 on the right. So let’s start by using 2 units of the adrenaline molecule on the left; that g ivesus18CO 2 ’s, and 13 H 2 O ’sonther igh t . And1N 2 product molecule. Without using any of the O 2 species, we have 6 oxygen units on the left, and require 18 × 2+13 × 1=49 oxygen atoms on the right. So we need 43 2 units of O 2 ; multiplying everything by 2 to get to integer coeﬃcients only (which is not strictly required) leads to: 4C 9 H 13 O 3 N + 43O 2 36CO 2 + 26H 2 O + 2N 2 2
2. (15 points) Trimethyl aluminum, Al(CH 3 ) 3 , must be handled carefully in an air-tight appa- ratus as it spontaneously bursts into ﬂame upon contact with oxygen. Calculate the mass of trimethyl aluminum which can be prepared from 5.00 g of aluminum metal and 25.0 g of dimethyl mercury via the chemical reaction 2Al(s) + 3Hg(CH 3 ) 2 (l) −→ 2Al(CH 3 ) 3 (l) + 3Hg(l) We need to decide which input is the limiting reagent as a start. That means fnding the number oF moles oF each reactant; 5.00 g oF Al metal with a molar mass 26.9815 corresponds to 0.1853 moles oF Al; 25.0 g oF dimethyl mercury has a molar mass oF (200 . 59 + 2 × (12 . 0112 + 3 × 1 . 00797)) = 230 . 66 g/mole; corresponding to 0.1084 moles. Our balanced equation states that we require three moles oF the dimethyl mercury reactant For each two moles oF Al; thus we have an excess oF Al metal, and our limiting reagent is the dimethyl mercury. Each mole yields two-thirds oF a mole oF our product; thereFore our maximum yield (in moles oF the trimethyl aluminum compound) is 0 . 1084 × 2 / 3=0 . 07226 moles oF trimethyl aluminum. The molar mass oF the trimethyl aluminum is 26 .

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## This test prep was uploaded on 02/15/2008 for the course CHEM 2090 taught by Professor Zax,d during the Fall '07 term at Cornell University (Engineering School).

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Prelim1ans_chem209_f - CHEMISTRY 209 PRELIM I October 4 2007 ANSWER KEY 1 1(a(7 points Balance the following chemical reaction B5 H9(g O2(g B2 O3(s

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