How many grams of dichloromethane result from reaction of 1.05 kg of methane if

Question

How many grams of dichloromethane result from reaction of 1.05 kg of methane if the yield is 42.7 % ? CH4Methane(g)+2Cl2Chlorine(g)CH2Cl2Dichloromethane(l)+2HCl(g) 1.15 g H2 is allowed to react with 9.98 g N2, producing 1.59 g NH3. Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units. SubmitHintsMy AnswersGive UpReview Part Part B What is the percent yield for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

CH4(g)+2Cl2(g)CH2Cl2(l)+2HCl(g)

from reaction,

1 mol CH4 = 2 mol Cl2 = 1 mol CH2Cl2

no of mol CH4 = 1.05*10^3/16 = 65.625 mol

theoretical yield of CH2Cl2 = 65.625*85 = 5578.125 grams

if 42.7% yield

mass of CH2Cl2 Produced = 5578.125*42.7/100 = 2381.86 grams

1.15 g H2 is allowed to react with 9.98 g N2, producing 1.59 g NH3

part A

no of mol of H2 = 1.15/2 = 0.575 mol

no of mol of N2 = 9.98/28 = 0.356 mol

limiting reagent is H2

theeoretical yield of NH3 = 0.575*(2/3)*17 = 6.517 grams

part B

percent yield = 1.59/6.517*100 = 24.4 %

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