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10.(20 polnts) Acrylonitrile, CoH,N is the starting material for the prnd (otilo

ID: 1024949 • Letter: 1

Question

10.(20 polnts) Acrylonitrile, CoH,N is the starting material for the prnd (otilowing Page 5 of 8 o synthetic fiber (acrylics). It can be produced acoran equation: production of a a. (10 points) How many grams of a crylonitrile can be produced if 51.0g of C3He reacts with 51.0g of nitrogenmonoxide. Answer: b. (6 points) How much (in grams) of the surplus reactant will be left over? Answer: c. (4 points) If the percentage yield of the reaction is typically 85.0%, how much product will be formed in an actual experiment? Answer:

Explanation / Answer

4C3H6 + 6 NO -------------- 4 C3H3N + 6 H2O + N2

mass of C3H6= 51.0 grams

molar mass of C3H6= 42 gram/mole

number of moles of C3H6 = 51.0/42= 1.214 moles

mass of NO= 51.0 grams

molar mass of NO= 30 gram/mole

number of moles of NO= 51.0/30 = 1.7 moles

according to equation

4 moles of C3H6 = 6 moles of NO

1.214 moles of C3H6 = ?

                                = 1.214x6/4= 1.821 moles of NO

we need 1.821 moles of NO. but we have 1.7 moles of NO . so NO is usedup first in the reaction.

Hence NO is limiting reagent.

according to equation

6 moles of NO= 4 moles of C3H3N

1.7 moles of NO= ?

                              = 4x1.7/6= 1.133 moles of C3H3N

number of moles of C3H3N formed = 1.133 moles

molar mass of C3H3N= 53 gram/mole

mass of 1.133 moles of C3H3N= 1.133 x53 = 60.049 grams

Theoritical yield of C3H3N= 60.05 grams.

B. C3H6 is excess reagent

6 mole of NO= 4 moles of C3H6

1.7 moles of NO= ?

                          = 4x1.7/6=1.133 moels of C3H6

number of moles of C3H6 reacted = 1.133 moles

number of moles of C3H6 remaining= 1.214 - 1.133= 0.081 moles

mass of 0.081 moles of C3H6= 0.081 x42= 3.402 grams

mass of surplus reactant = 3.402 grams

C) percent yield= 85.0%

Theoritical yield= 60.05 grams

Percent yield= Actual yiels/Theoritical yield x100

Actual yield= 85.0x60.05/100 = 51.04 grma

Actual yield= 51.04 grams.

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