class6 - Class #6 Solution Stoichiometry CHEM 107 A....

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Class #6 Solution Stoichiometry CHEM 107 A. Bengali Texas A&M University Qatar
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Reaction Stoichiometry ) Same idea as for pure substances ) Balanced equation! ) Moles! ) Use concentrations and volumes to find numbers of moles
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Titration Example ) Titration: find concentration of solution by reaction H 2 SO 4 + 2NaOH Na 2 SO 4 + 2 H 2 O ) If 20.00 mL of NaOH neutralizes 25.00 mL of 0.0800 M H 2 SO 4 , what is the NaOH concentration?
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Titration H 2 SO 4 + 2NaOH Na 2 SO 4 + 2 H 2 O H 2 SO 4 : 25 mL, 0.08 M NaOH: 20 mL, ?? M Moles H 2 SO 4 ? 0.025 L x 0.08 M = 0.002 mol
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Titration H 2 SO 4 + 2NaOH ... 0.002 mol ? mol 0.002 mol H 2 SO 4 x 2molNaOH 1mol H 2 SO 4 = 0.004 mol NaOH Now use moles and volume to find molarity
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Titration NaOH: 0.004 mol in 20 mL 0.004 molNaOH 0.020L = 0.20 M NaOH
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) Same idea as before! ) Identify limiting reagent, then solve problem ) Mg in HCl experiment: 1.5 g Mg reacts with 30 mL of 2.0 M HCl. What’s in the flask after reaction?
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class6 - Class #6 Solution Stoichiometry CHEM 107 A....

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