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4. Consides the reactions etely + de cely Hel . If 11.99 e Hel with excers Chlosine produced 11.og eely Whatń the % field? an
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Answer #1

4) 11.9 g CHCl3 = mass / molar mass = 11.9 g / 119.38 g/mole = 0.09968 mole.

one mole CHCl3 react with one mole Cl2 to produce one mole CCl4.

thus

theoretical yield of CCl4 = 0.09968 mole * 153.82 g/mole = 15.33 g

% of yield = 11.0 * 100 / 15.33 = 71.7 %

5)

7.2 g NaCl = mass / molar mass = 7.2 g / 58.5 g / mole = 0.1231 mole.

molarity = 0.1231 mole / 0.100 L = 1.23 M

6)

256 ml of 0.875 M NH4Cl = 0.256 L * 0.875 mole / L = 0.224 mole.

mass of NH4Cl = mole * molar mass = 0.224 mole * 53.491 g/mole = 11.98 g

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