Problem Set 1_Sol

# Problem Set 1_Sol - Answer Key Ch 1a, Problem Set One Side...

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Unformatted text preview: Answer Key Ch 1a, Problem Set One Side 1 of 10 Problem One Stochiometry 20 Points a. 2N 2 H 4 + N 2 O 4 &amp;quot; 3N 2 + 4H 2 O (1) CH 3 ( ) 2 N 2 H 2 + 2N 2 O 4 &amp;quot; 3N 2 + 4H 2 O+2CO 2 (2) b. The mixture is 50/50 by mass hydrazine and dimethylhydrazine. Therefore if the total mass is 2186 kg, each component is 1093 kg (since each fraction is known to infinite precision). Assuming that all of the reactants are used up, we can use this mass, the molar masses of each reactant, and the stoichiometric ratios from the above balanced equation to determine the mass of nitrogen tetroxide Hydrazine: 1651.5x10 3 g N 2 H 4 1mol N 2 H 4 32.05g N 2 H 4 &amp;quot; # \$ % &amp;amp; 1mol N 2 O 4 2mol N 2 H 4 &amp;quot; # \$ % &amp;amp; = 25764mol N 2 O 4 Dimethylhydrazine 1651x10 3 g CH 3 ( ) 2 N 2 H 2 1mol CH 3 ( ) 2 N 2 H 2 60.10g CH 3 ( ) 2 N 2 H 2 &amp;quot; # \$ \$ % &amp;amp; 2mol N 2 O 4 1mol CH 3 ( ) 2 N 2 H 2 &amp;quot; # \$ \$ % &amp;amp; = 54959mol N 2 O 4 So, 80723 total moles N 2 O 4 are used. Using the molar mass of N 2 O 4 (=92.01 g/mol), the mass can be found to be 7427.3 kg, or with the correct number of sig figs Final Answer: 7.427 x10 3 kg N 2 O 4 used The stoichiometric ratio of carbon dioxide is the same for nitrogen tetroxide in the decomposition of dimethylhydrazine, so the same molar amount will be produced 1651.5x10 3 g CH 3 ( ) 2 N 2 H 2 1mol CH 3 ( ) 2 N 2 H 2 60.10g CH 3 ( ) 2 N 2 H 2 &amp;quot; # \$ \$ % &amp;amp; 2mol CO 2 1mol CH 3 ( ) 2 N 2 H 2 &amp;quot; # \$ \$ % &amp;amp; 44.01g CO 2 1mol CO 2 &amp;quot; # \$ % &amp;amp; = 2419kg C Final Answer: 2.419 x10 3 kg CO 2 evolved 2.5 points each -0.5 for each wrong coefficient 2 points each equation 2 points, -1 for incorrect number of sig figs 2 points 2 points, -1 for incorrect number of sig figs Answer Key Ch 1a, Problem Set One Side 2 of 10 c. First determine the molar amount of water produced by this reaction. 1651.5x10 3 g N 2 H 4 1mol N 2 H 4 32.05g N 2 H 4 &amp;quot; # \$ % &amp;amp; 2mol H 2 O 1mol N 2 H 4 &amp;quot; # \$ % &amp;amp; = 103057mol H 2 O 1651.5x10 3 g CH 3 ( ) N 2 H 2 1mol CH 3 ( ) N 2 H 2 60.10g CH 3 ( ) N 2 H 2 &amp;quot; # \$ % &amp;amp; 4mol H 2 O 1mol CH 3 ( ) N 2 H 2 &amp;quot; # \$ % &amp;amp; = 109826mol H 2 O Or 221883 mol H 2 O total. There are molar mass of water is 18.02 and the density is given as 0.92 g/cm 3 . Putting this all together into a cylinder 1.0 km in radius: h = mol H 2 O &amp;quot; MW # r 2 \$ = 140891mol H 2 O ( ) &amp;quot; 18.01g/mol ( ) # &amp;quot; 10 5 cm ( ) 2 &amp;quot; 0.92g/cm 3 ( ) Final Answer = 1.3 x 10-4 cm 1 point for each equation 1 point 2 points, -1 for sig figs Answer Key Ch 1a, Problem Set One Side 2 of 10 c. First determine the molar amount of water produced by this reaction. 1651.5x10 3 g N 2 H 4 1mol N 2 H 4 32.05g N 2 H 4 &amp;quot; # \$ % &amp;amp; 2mol H 2 O 1mol N 2 H 4 &amp;quot; # \$ % &amp;amp; = 103057mol H 2 O 1651.5x10 3 g CH 3 ( ) N 2 H 2 1mol CH 3 ( ) N 2 H 2 60.10g CH 3 ( ) N 2 H 2 &amp;quot; # \$ % &amp;amp; 4mol H 2 O 1mol CH 3 ( ) N 2 H 2 &amp;quot; # \$ % &amp;amp; = 109826mol H 2 O Or 221883 mol H 2 O total. There are molar mass of water is 18.02 and the density is given as 0.92 g/cm 3 . Putting this all together into a cylinder 1.0 km in radius: ....
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## This note was uploaded on 03/30/2010 for the course CH 1a taught by Professor Lewis during the Fall '08 term at Caltech.

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Problem Set 1_Sol - Answer Key Ch 1a, Problem Set One Side...

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