Problems: 2) 3) Nitrogen and hydrogen can combine (in the presence of a catalyst

Question

Problems:

2)

3)

Nitrogen and hydrogen can combine (in the presence of a catalyst) to produce ammonia according to the equation below: N2(g) + 3 H2(g) rightarrow NH3(g) If 4 molecules of nitrogen and 9 molecules of hydrogen are present in a reaction vessel, draw a picture representing the molecules found in the vessel before and after the reaction: How many molecules of ammonia are produced? What was the limiting reactant? How many molecules of the excess reactant remain? What is the maximum mass of S8 that can be produced from the following reaction if 85.0g of each reactant were combined? 8 SO2 + 16 H2S rightarrow 3 S8 + 16 H2O Use the following chemical equation to answer the following questions: 2 H2O2(l) + N2H4(I) rightarrow 4 H2O(g) + N2(g) Calculate the mass (in grams) of nitrogen gas produced when 45.0g of hydrogen peroxide is combined with 25.0g of N2H4. Calculate the mass (in grams) of excess reactant remaining after the reaction goes to completion. Calculate the mass (in grams) additional limiting reactant needed to use up the excess reactant.

Explanation / Answer

problem#1

N2 + 3H2 ==>2NH3

from balance reaction

1molecule of N2 stiochiometrically reacts with=3molecules of H2 hence 4molecules of N2 will require 12molecules of H2 while available H2 is 9molecules. therefore H2 is a limiting reagent. so

3 molecules(N2) + 9molecules(H2)===>6molecules of NH3

a) 6

b) H2

c) 1molecule of N2 excess

problem#2

8SO2 + 16H2S==>3S8 + 16H2O

SO2 + 2H2S ==>3/8S8 + 2H2O

64g(SO2) + 68g(H2S) ===>96g(S8) + 36g(H2O)

since 85 grams of each reactant is present therefore H2S is in excess so divide 0.8

80g(SO2) +85g(H2S) ===>120g(S8) + 45g(H2O)

the maximum mass of S8 that cab be produced is=120g

Problem#3

2H2O2 + N2H4 ==> 4H2O + N2

68g(H2O2) + 32g(N2H4) ==> 72g(H2O) + 28g(N2) , since H2O2 is limiting reagent so divide all terms by 1.5111

45g(H2O2) + 21.18g(N2H4) ==>47.65g(H2O) + 18.53g(N2)

a) mass of N2 gas produced=18.53g

b) mass of excess N2H4=25-21.18=3.82g N2H4

c) mass of additional limiting reagent to consume excess reagent

21.18g N2H4 reacts stiochiometrically with=45g H2O2

3.82g N2H4 reacts stiochiometrically with=45/21.18*3.82g H2O2=8.116g H2O2

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