H01 Stoichiometry-solutions

H01 - nanni(arn437 H01 Stoichiometry vanden bout(50985 This print-out should have 16 questions Multiple-choice questions may continue on the next

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nanni (arn437) – H01 Stoichiometry – vanden bout – (50985) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Which compound has the wrong chemical For- mula? 1. CaOH correct 2. Ba 3 (PO 4 ) 2 3. (NH 4 ) 2 SO 4 4. Mg(OH) 2 Explanation: The calcium ion is Ca 2+ ; the hydroxide ion is OH . Two OH are needed to balance the charge oF each Ca 2+ , so the Formula is Ca(OH) 2 . The magnesium ion is Mg 2+ ; the hydroxide ion is OH . Two OH are needed to balance the charge oF each Mg 2+ , so the Formula is Mg(OH) 2 . The barium ion is Ba 2+ ; the phosphate ion is PO 3 4 . Two PO 3 4 are needed to balance the charge oF every three Ba 2+ . (This gives a total anion charge oF - 6 and a total cation charge oF +6.) The Formula is Ba 3 (PO 4 ) 2 . The ammonium ion is NH + 4 ; the sulFate ion is SO 2 4 . Two NH + 4 are needed to balance the charge oF each SO 2 4 , so the Formula is (NH 4 ) 2 SO 4 . 002 10.0 points Which one has the greatest number oF atoms? 1. 3.05 moles oF helium 2. 3.05 moles oF water 3. 3.05 moles oF CH 4 correct 4. 3.05 moles oF argon 5. All have the same number oF atoms Explanation: ±or 3.05 moles oF water: ? atoms = 3 . 05 mol H 2 O × 6 . 02 × 10 23 molec 1 mol × 3 atoms 1 molecule = 5 . 51 × 10 24 atoms ±or 3.05 moles oF CH 4 : ? atoms = 3 . 05 mol CH 4 × 6 . 02 × 10 23 molec 1 mol × 5 atoms 1 molecule = 9 . 18 × 10 24 atoms ±or 3.05 moles oF helium: ? atoms = 3 . 05 mol He × 6 . 02 × 10 23 atoms 1 mol = 1 . 84 × 10 24 atoms ±or 3.5 moles oF argon: ? atoms = 3 . 05 mol Ar × 6 . 02 × 10 23 atoms 1 mol = 1 . 84 × 10 24 atoms 003 10.0 points IF 100.0 grams oF copper (Cu) completely re- acts with 25.0 grams oF oxygen, how much copper(II) oxide (CuO) will Form From 140.0 grams oF copper and excess oxygen? ( Note : CuO is the only product oF this reaction.) 1. 175.0 g correct 2. 200.0 g 3. 160.0 g 4. 150.0 g 5. 35.0 g Explanation: m Cu , ini = 100.0 g m O 2 = 25.0 g m Cu , fn = 140.0 g IF 100 g copper and 25 g oxygen react com- pletely with each other, there must be 125 g oF product Formed (law oF conservation oF mass). This product is CuO.
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nanni (arn437) – H01 Stoichiometry – vanden bout – (50985) 2 Now we have a ratio: for every 100 g of Cu reacted, 125 g of CuO will be produced (assuming there is enough oxygen). We use this ratio to Fnd the mass of CuO that could be formed from 140 g of Cu and excess oxygen. We set our known ratio (100 g Cu : 125 g
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This note was uploaded on 09/19/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.

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H01 - nanni(arn437 H01 Stoichiometry vanden bout(50985 This print-out should have 16 questions Multiple-choice questions may continue on the next

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