solutions - Version 020 Exam 2 McCord (91750) This...

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Version 020 – Exam 2 – McCord – (91750) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Some sulfdes are converted to oxides by roast- ing (heating in air). 2 ZnS + 3 O 2 + heat 2 ZnO + 2 SO 2 IF 2.5 moles oF zinc sulfde are heated with 4.0 moles oF oxygen, which reactant is in ex- cess, and by how much? 1. O 2 ; by 1.0 moles 2. ZnS; by 0.17 moles 3. O 2 ; by 0.25 moles correct 4. ZnS; by 0.50 moles Explanation: n ZnS = 2 . 5 mol n O 2 = 4 . 0 moles We recognize this as a limiting reactant problem because the amounts oF more than one reactant are given. We must determine which oF these would be used up frst (the limiting reactant). To do this we compare the required ratio oF reactants to the available ratio oF reactants. The balanced chemical equation shows that we need 2 mol ZnS For every 3 mol O 2 . We use these coe±cients to calculate the required ratio oF reactants: 2 mol ZnS 3 mol O 2 = 0 . 667 mol ZnS 1 mol O 2 ²rom this ratio we see that each mole oF O 2 that reacts requires exactly 0.667 moles oF ZnS. Next we calculate the available ratio oF reactants From our data: 2 . 5 mol ZnS 4 . 0 mol O 2 = 0 . 625 mol ZnS 1 mol O 2 We have only 0.625 mol ZnS available For each mole oF O 2 , not enough ZnS to react all oF the O 2 . We will run out oF ZnS frst, so ZnS is the limiting reactant. There will be some O 2 leFt unreacted or in excess. To calculate how much O 2 will be leFt over, we must frst calculate how much O 2 is used up in reacting all the ZnS: ? mol O 2 reacted = 2 . 5 mol ZnS × 3 mol O 2 2 mol ZnS = 3 . 75 mol O 2 We fnd the amount oF excess O 2 by sub- tracting the amount oF O 2 used up From the starting amount oF O 2 : ? mol O 2 leFt over = 4 . 0 mol O 2 - 3 . 75 mol O 2 = 0 . 25 mol O 2 002 9.0 points What hybridization would you expect For Se when it is Found in SeO 2 - 4 ? 1. s 2 p 2 2. p 3. sp 4. sp 3 d 5. sp 2 6. spd 7. sp 3 d 2 8. sp 3 correct Explanation: N = 5 × 8 = 40 A = (6 × 5) + 2 = 32 S = 40 - 32 = 8 There are 8 shared or bonded electrons. The remaining electrons are placed around the O atoms. There are no lone pairs on the
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Version 020 – Exam 2 – McCord – (91750) 2 central Se atoms. This molecule has tetra- hedral electronic and molecular geometry so there are 4 regions of high electron density. This corresponds to sp 3 hybridization. 003 9.0 points What is the shape of CS 2 - 3 ? 1. tetrahedral 2. trigonal planar correct 3. trigonal pyramidal 4. seesaw 5. T-shaped Explanation: There are three regions of electron density (with no lone pairs) around the central atom: C S b S S b b 2 - 004 9.0 points What is the molar mass of a compound if 75.5 g of the compound dissolved in 2.05 L gives a molarity of 0.555 M? 1.
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solutions - Version 020 Exam 2 McCord (91750) This...

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