1)It is desired to produce 27.8 grams of iron(II) oxide by the following reactio

Question

1)It is desired to produce27.8grams ofiron(II) oxideby the following reaction. If the percent yield ofiron(II) oxideis91.4%, how many grams ofoxygen gaswould need to be reacted?

2)An iron nail rusts when exposed to oxygen.For the following reaction,3.09grams ofoxygen gasare mixed with excessiron. Assume that the percent yield ofiron(III) oxideis65.5%.

3)

For the following reaction,6.68grams ofhydrogen gasare mixed with excessnitrogen monoxide. Assume that the percent yield ofnitrogen gasis84.4%.

Explanation / Answer

2Fe + O2 ---->2 FeO , 27.8 gm FeO = 27.8/71.844 = 0.387 moles,

hence number of moles of oxygen = (0.387/2) = 0.19347 ,

Oxygen in grams required considering 91.4 % process = ( 0.19347x32)x(100/91.4) = 6.77 grams ,

part 2) oxygen moles = 3.5/32 =0.10937 ,

4Fe + 3O2 ---> 2Fe2O3 , moles of Fe2O3 = 0.10937 x2/3 = 0.0729 ,

theoretical yiesld = ( 0.0729 x 159.69) = 11.64,

actual yield = ( 11.64 x65.5/100) = 7.627 gramsn,

part 3) 2NO +2 H2 ---> N2 +2 H2O , moles of Hydrogen = 6.68/2 = 3.34,

nitrogen gas moles = 3.34/2 = 1.67 ,

theoretical yield = 1.67 x 30 = 50.1 gm ,

actual yield = (50.1 x84.4/100) = 42.28 grams

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