chemistry question on limiting reagents

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153 grams of ammonia and 325 grams of oxygen were reacted according to the following equation:

4NH3 + 5O2 --> 4NO + 6H2O

a) what is the limiting reagent?

b) how many grams of each are produced?

c) what is the percent yield assuming that 204 grams of water were actually produced?

d) compare the total masses of reactant to the total mass of products. what law does this prove?

Apr 26th, 2015

4NH3     +   5O2  ------        4NO         +    6 H2O

 9.0 mol     10.16 mol         8.13 mol           12.19 mol

 a)  4 moles of NH3 react with 5 moles of O2 hens 

          9.0 moles of NH3 required 5/4*9 moles of O2 = 11.25 moles of O2 is required 

 But there are 10.16 moles of O2 is present in reaction flask  hence 

Limiting reagent will be O2


B) NO = 8.13 x 30 = 243.9 g , H2O = 12.19 x 18 = 219.46 g

C) % yield = 100 x 204/ 219.6 = 92.89 %

D) Mas of reactant(478g)  , mass of product( 243.9+219.46 = 463.36)

 There should be mass of reactant = mass of product by law of conservation of mass 

 but NH3 is not reacting completely  so  it is not follow this law.


Apr 28th, 2015

Regarding any doubt please let me know I will be happy to help you.

Apr 28th, 2015

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