24.0 g of nitrogen gas react with an excess of hydrogen gas to give an actual yi

Question

24.0 g of nitrogen gas react with an excess of hydrogen gas to give an actual yield of 3.85 g NH3. What is the percent yield for this reaction? N2(g)+3H2(g)-->2NH3(g)
24.0 g of nitrogen gas react with an excess of hydrogen gas to give an actual yield of 3.85 g NH3. What is the percent yield for this reaction? N2(g)+3H2(g)-->2NH3(g)
24.0 g of nitrogen gas react with an excess of hydrogen gas to give an actual yield of 3.85 g NH3. What is the percent yield for this reaction? N2(g)+3H2(g)-->2NH3(g)

Explanation / Answer

we cans ee that 1 mole of N2 gives 2 mols of NH3

or 24/28 mols of nitrogen will give ..... x moles

x moloes = (24/28) X 2 = 1.71 mols (molecular weight of nitrogen = 28)

or moles theoritical yield = 1.71 X 17 = 29.1 g (molecular weight of NH3 = 17)

and practical yield = 3.85g

and we know that % yield = (practical yield / theoretical yield) X 100

= (3.85/29.1) X 100

13.23 %

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