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post-lab 5 ch 204

# post-lab 5 ch 204 - obennoskey(blo242 – Post-lab 5 –...

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Unformatted text preview: obennoskey (blo242) – Post-lab 5 – Lyon – (52635) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 250.0 ml of 0.460 M nickel chlorate solution is reacted with 130.0 ml of 1.60 M potassium hy- droxide. The unbalanced molecular equation for this reaction is: Ni(ClO 3 ) 2 + KOH → Ni(OH) 2 + KClO 3 Balance the equation for this reaction and determine the limiting reagent for this reac- tion using the information given above. Then report the number of moles of Ni(OH) 2 that can be formed before the limiting reagent runs out. Correct answer: 0 . 104 moles. Explanation: There are many different ways to do lim- iting reagent problems. This is just one way you could do it. First you need a balanced equation: Ni(ClO 3 ) 2 + 2 KOH → Ni(OH) 2 + 2 KClO 3 . 2500 L × . 460 M = 0 . 115 moles Ni . 1300 L × 1 . 60 M = 0 . 208 moles OH- Since you need 1 Ni for every Ni(OH) 2 , you have enough nickel to make 0.115 moles of product. Since you need 2 OH- for every Ni(OH) 2 , you have enough hydroxide to make 0.104 moles of product. You will therefore run out of hydroxide before you run out of nickel. Hydroxide is the limiting reagent, and the amount of product formed will be 0.104 moles. 002 10.0 points Determine the limiting reactant for the fol- lowing reaction, given that 100 grams of each reactant was used. 3 Cu (s) + 2 H 2 O(l) + SO 2 (g) + 2 O 2 (g) → Cu 3 (OH) 4 SO 4 (s) 1. Cu (s) - 0.525 moles of product formed correct 2. SO 2 (g) - 0.525 moles of product formed 3. H 2 O (l) - 0.525 moles of product formed 4. SO 2 (g) - 1.56 moles of product formed 5. Cu (s) - 1.56 moles of product formed 6. O 2 (g) - 2.77 moles of product formed 7. O 2 (g) - 1.56 moles of product formed 8. H 2 O (l) - 2.77 moles of product formed Explanation: There are many different ways to do lim- iting reagent problems. This is just one way you could do it. 1) 100 grams of copper metal divided by its atomic weight (63.546 g/mole) gives you moles of copper. According to the balanced equation, you will get 1 mole of Cu 3 (OH) 4 SO 4 for every 3 moles of Cu, so the moles of Cu 3 (OH) 4 SO 4 formed is equal to 1/3 the moles of Cu. If you do the math, 100 g of Cu is enough to form 0.525 moles of product....
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post-lab 5 ch 204 - obennoskey(blo242 – Post-lab 5 –...

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