Hydrogen gas is generated in this reaction by reaction of Pd with NH_4^+ HCO_2^-

Question

Hydrogen gas is generated in this reaction by reaction of Pd with NH_4^+ HCO_2^- to give CH_4. N_2 and H_2 reaction of Pd with NH_4^+ HCO_2^- to give NH_3, CO_2 and H_2 reaction of Pd with NN_4^+ HCO_2^- to give to N_2, CO_2 and H_2 reaction of Pd with CH_3OH to give CO_2 and H_2 In the alkene hydrogenation reaction shown above, palladium serves as a catalyst the limiting reagent an inert support both a and b Which of the following reactions does the alkene undergo in the above reaction? elimination oxidation reduction C=C bond cleavage If you were to use 0.59 g of the alkene reactant, how many grams of ammonium formate is required if a 4.0 mole equivalent of ammonium format (MW= 63.06 g/mol) relative to the alkene is needed for the reaction? 2.7 g 0.94 g 1.1 g 0.68 g After completion of the above reaction and recrystallization of the product (B), a permanganate test was performed. A negative permanganate test was observed, indicating that The reaction did not successfully produce B The product contained larger amounts of A than of B The reaction was complete and no A was present The permanganate test was not relevant for this experiment

Explanation / Answer

Q23

alkene -- > alkane, must be reduction, since addition of Hydrogen is typically decrease in oxidation number

Q24

m = 0.59 g of reactant...

find ammonium formate required

n = 4 mol equiv....

mol alkene= mass/MW = 0.59/218.33 = 0.002702 mol

so we require 4:1 = 4*0.002702 = 0.010808 mol of of ammonium formate

so

mass = mol*MW = 0.010808*63.06 = 0.6815524 g choose D

Q25

if no permanganate is shown, there is no alkene, so w ehave plenty of product

choose C

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