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mass of magnesium
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- Words:
- 1403
- Submitted:
- Wed Apr 23 2008
- Mark submitted by Author:
... DETERMINATION OF RAM OF MAGNESIUM Method1 Mass of Mg = 0.12g Volume of Hydrogen = 128cm3 Treatment of results Mg + 2HCl ==> MgCl2 +H2 * n= v?V ==> 128 ?24000 = 5.33?10-( moles * mole ratio is 1:1, therefore moles of mg that reacted= 5.33?10-( moles * RAM of Mg (n= m?M ==> M= m?n) ==> 0.12 ? 5.33?10-( = 22.51 Method2 Mass of magnesium from method 1 = 0.12 Weight of boat = 44.17 Weight of boat with solution = 52.84 Weight of boat with salt (MgCl2) = 44.74 o Mass of MgCl2 formed = 44.74 ? 44.17 = 0.57 o Mass of Cl- ions = 0.57 ? 0.12 = 0.45g o n= m ?M = 0.45 ? 35.5 = 0.013moles o n= m ?M = 0.57 ? 24.3 = 0.023moles EVALUATION Method Result 1 22.5 2 18.9 Expected 24.3 From looking at the end results in both methods, I can clearly see that method 1 was the more accurate and appropriate way to determine the mass
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