Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 10 Gases and Their Properties - Study Questions - Page 403h: 103


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Work Step by Step

1) $d=m/V=PM/RT$ $\bar{M}=0.195\ g×0.082×308\ K/((725/760×1\ atm)(0.125\ L))$ $\bar{M}=41.33\ g/mol$ 2) Let x be the mol fraction of $NO_2$: $x×46.00+(1-x)×32.00=41.33\ g/mol$ $x=0.666=66.6\%$ $O_2:0.334=33.4\%$ 3) Number of moles: $NO_2: 0.195\ g/41.33\ g/mol×0.666=3.142×10^{-3}\ mol$ $O_2: 0.195\ g/41.33\ g/mol×0.334=1.576×10^{-3}\ mol$ 4) The ratio of the number of moles of the two is: $2.00$ Stoichiometry gives: $Cu(NO_3)_2\rightarrow CuO+1/2\ O_2+2\ NO_2$ Stoichiometric ratio: 4.0 Since $N_2O_4$ is liquid at this Temperature and Pressure, the difference in ratios could be explained by the formation of the tetroxide.
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