Chem 161-2009 Homework 4th week

# Chem 161-2009 Homework 4th week - Chem 161-2009 Homework...

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Chem 161-2009 Homework 4 th Week 14. For some applications, solution concentrations are expressed as grams of solute per liter of solution, and for some, as milligrams of solute per milliliter of solution. How are these two concentration units related? Explain. They are exactly the same. A gram is 1000 mg. A liter is 1000 mL. Therefore, 1 g of solute per L of solution = 1000 mg of solute per 1000 mL of solution. This simplifies to mg of solute per mL of solution. 75. Liquid mercury and oxygen gas react to form mercury(II) oxide. Without doing detailed calculations, decide which of the following should result from the reaction of 0.200 mol Hg(l) and 4.00 g O 2 (g). (a) 4.00 g HgO(s) and 0.200 mol Hg(l); (b) 0.100 mol HgO(s), 0.100 mol Hg(l), and 2.40 g O 2 (g); (c) 0.200 mol HgO(s) and no O 2 (g); or (d) 0.200 mol HgO(s) and 0.80 g O 2 (g)? Explain. Hg(l) + O 2 (g) → HgO 2 Hg(l) + O 2 (g) →2 HgO AW Hg = 200.59 AW O 2 = 32.00 AW HgO = 200.59 + 16.00 = 216.59 4.00 g O 2 = 4.00g/32.00 gmol -1 = 0.125 mol O 2 . Since the stoichiometric ratio of Hg to O 2 is 2:1, and the actual ratio is 0.200 to 0.125, then the Hg is the limiting reactant, and the O 2 is in excess. Hence, all of the Hg should become HgO (theoretically), leaving some O 2 left over. (a) is incorrect because it shows Hg left over, not O 2 . (b) shows that most of the O 2 is left over, which doesn’t make sense since the O 2 is only in slight excess. (c) shows no O 2 left over, which is incorrect, because the O 2 is in excess. (d) shows a lot of HgO formed, and only a small amount of O 2 left over, which makes sense. 83. A student prepares ammonium bicarbonate by the reaction NH 3 + CO 2 + H 2 O → NH 4 HCO 3 She uses 14.8 g NH 3 and 41.3 g CO 2 . Water is present in excess. What is her actual yield of ammonium bicarbonate if she obtains a 74.7% yield in the reaction? MW NH 3 = 14.01 + 3.03 = 17.04 MW CO 2 = 12.01 + 32.00 = 44.01 MW NH 4 HCO 3 = 14.01 + 4.04 + 1.01 + 12.01 + 48.00 = 79.07 moles of NH 3 = 14.8 g/17.04 gmol -1 = 0.869 mol moles of CO 2 = 41.3 g/44.01 gmol -1 = 0.938 mol Hence, both water and CO 2 are in excess. Theoretical yield of NH 4 HCO 3 = 0.869 mol x 79.07 gmol -1 = 68.71 g NH 4 HCO 3 Since she only gets 74.7% of the theoretical yield, her actual yield = 68.71 x 0.747 = 51.33g NH 4 HCO 3 . 1

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85. Calculate the molarity of each of the following aqueous solutions. (a) 2.60 mol CaCl 2 in 1.15 L of solution (b) 0.000700 mol Li 2 CO 3 in 10.0 mL of solution (c) 6.631 g NaNO 3 in 100.0 mL of solution MW NaNO 3 = 22.99 + 14.01 + 48.00 = 85 (d) 412 g sucrose, C 12 H 22 O 11 , in 1.25 L of solution. MW C 12 H 22 O 11 = (12 x 12.01) + (22 x 1.01) + (11 x 16.00) = 342.34 (e) 15.50 mL glycerol, C 3 H 8 O 3 (d = 1.265 g/mL) in 225.0 mL of solution MW C 3 H 8 O 3 = (3 x 12.01) + (8 x 1.01) + (3 x 16.00) = 92.11 (f) 35.0 mL 2-propanol, CH 3 CHOHCH 3 , (d = 0.786 g/mL) in 250 mL of solution. MW CH
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Chem 161-2009 Homework 4th week - Chem 161-2009 Homework...

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