By Martin Silberberg

Through Patricia Amateis of Virginia Tech. This complement comprises distinctive suggestions and reasons for all even-numbered difficulties typically textual content. .

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Extra resources for Student Solutions Manual to accompany Principles of General Chemistry

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340 mol AICI) Ammonium chloride is the limiting reactant, and it is important for all subsequent calculations. 49 g NH4CI 3 mol NH4CI I mol AI(N02 )3 = 5 6 . 1 4474 = 56. 1 g AI(N02») AI(N02) 3 remaining: 62 . 5 g - 56. 4 g A I(N 02 h )( ( )( ) NH4CI: none left since it is the limiting reagent. 02 J J 1 33 . 4 g A ICI3 = 2 8 . 6 g N 2 Plan: Multiply the yield of the first step by that of the second step to get the overall yield. Solution : It is simpler to use the decimal equivalents of the percent yields, and then convert to percent using 1 00%.

Either the lead or the chloride ion is limiting - the one producing the fewer moles of product is limiting. Convert the lower number of moles of product to grams of product. Solution: Balanced reaction: Pb(C 2 H 3 0 2 Maq) + N aCI(aq) � PbCI 2 (s) + NaC 2 H 3 0 2 (aq) Because of the l : l ratio, there will be mol of PbCI 2 produced from mol of Pb 2+. The moles of PbCI 2 produced from the NaCI are: 1 O -3 mol NaCI --I mol PbCI 2 = mol PbCI 2 mol NaCI Im L The NaCI is limiting. The mass of PbCI 2 may now be determined using the molar mass.

00) = 1 8 7 . 4 1 mol C U 2 C03 ) ( 1 8 7 . 1 1 g Cu 2 CO 3 I mol C U 2 C03 ) = 1 5 7 3 . 57 x 3 1 0 g CU 2 C03 b ) D i n i trogen pentaox ide has the formula N20S ' Di- indicates 2 N atoms and penta- i ndicates 5 ° atoms. M of N20s = ( 2 x 1 4 . 00) = 1 08 . 02 gimol M ass N20S = ( 2 . 04 X 10 21 N 2 0S mOlecu les = 0 . 366 g N 2 0S ) [ 34 1 �3O l N 2 0S 6 . 02 2 x 1 0 N 2 0S molecules J( 1 08 . 02 g N 2 0S 1 mol N 2 0S ) c) The correct formula for this ionic compound is NaCI04. There are Avogadro ' s number of entities (in this case, formula units) in a mole of this compound .

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