4 and 5 please 4. Reaction Scheme- Fill in the Blanks (3 points): H2 10% Pd/C OM

Question

4 and 5 please

4. Reaction Scheme- Fill in the Blanks (3 points): H2 10% Pd/C OMe OMe MeOH H2 Pd/C MeOH 10% MW or Conc. 296.49 2.02 volume 0.7 mL 2L 3.0 mL density mass mmole equivalents Theoretical Yiels 350 mg 15 mg Theoretical Meld catalyst solvent * The methyl oleate used in this experiment is only 70% pure 0.5 g of liquid only contains 0.35 g of methyl oleate) *22.4 L/mole gas@ STP. Assume STP 5. a) Show your calculations for determining the maximum possible amount (- maximum theoretical yiel methyl stearate that could be formed in the reaction above. (1 point): b) What is the percentage yield if 237 mg of methyl stearate is produced from the reaction above (Show calculations) (1 point)

Explanation / Answer

5a) theoretical yield : reactant (mol weight = 296.49 g) , product (molecular weight = 298.49) as the product just have two extra hydrogens.

296.49 g reactant gives 298.49 g product

1 g reactant gives 298.49/296.49 = 1.006 g product

350 mg or 0.35 g reactant gives 0.35 * 1.006 = 0.352 g product(100% yield)

352 mg is the theoretical 100 % yield

5b) percentage yield = (experimental yield/theoretical yield ) * 100

= 0.237/0.352 * 100 = 67.3 %

4) MeOH mole cular weight = 32.04 g/mol ; density = 792 kg/m3 ; mmol = mass/molar mass * 10-3

MeOH -mmol = 74 mmol

methyl stearate density = 0.9 g/mL

H2 density = 0.089 g/L 0; mass = 0.178 g

  

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