Takenote11 - Solution 1 C12 P4 —> PC13(0.337 moles...

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Unformatted text preview: Solution: 1. C12 + P4 —-> PC13 (0.337 moles PC13) (1 mole P4_) (123.8953 g P4) 4 moles PC13 1 mole P4 =10.44 gP4 2. Similarly for C1; = 35.84 g C12 B. Some important points. (Covered in Chapter 4) 1. Limiting Reagents 2. Theoretical Yield 3. Actual Yield. 4. %Yield. C. Q.) 9.21 g CH3NHNH3 reacts with 32 gms of 02 Which reactant is limiting? Solution: 1. CH6N2 —> C03 ‘1' 31—120 ‘1' N2 2. (9.21 g CHfiNg) (1 mol CHéfi) 46 g CH6N2 = 0.20 moles CH6N2 3. (32 g 02) (1 mo] 01) 32 g 02 = 1 mol 02 Thus, for every 0.20 moles of CH6N2 used, 1 mole 02 is required. Hence CH6N2 is the limiting reactant....
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