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prisoha [69]
3 years ago
7

In one experiment 7.62 g of Fe are allowed to react with 8.67 g of S

Chemistry
1 answer:
vichka [17]3 years ago
4 0

calculate moles of both reagents given and the moles of FeS that each of them would form if they were in excess  

moles = mass / molar mass  

moles Fe = 7.62 g / 55.85 g/mol  

= 0.1364 moles  

1 mole Fe produces 1 mole FeS  

Therefore 7.62 g Fe can form 0.1364 moles FeS  

moles S = 8.67 g / 32.07 g/mol  

= 0.2703 moles S  

1 mole S can from 1 moles FeS  

So 8.67 g S can produce 0.2703 moles FeS  

The limiting reagent is the one that produces the least product. So Fe is limiting.  

The maximum amount of FeS possible is from complete reaction of all the limiting reagent.  

We have already determined that the Fe can form up to 0.1364 moles of FeS, so this is max amount of FeS you can get.  

Convert to mass

hope this helps :)

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The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane wi
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Explanation:

mass fraction N₂ : He : CH₄ : C₂H₆ : : 15 : 5 : 60 : 20

mole fraction  N₂ : He : CH₄ : C₂H₆ : : 15/28 : 5/4 : 60/16 : 20/30

mole fraction  N₂ : He : CH₄ : C₂H₆ : : .5357  : 1.25 : 3.75 : .67

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Partial pressure = total pressure x mole fraction

Partial pressure of N₂ = 1200 kPa  x .0877 = 105.24 kPa

Partial pressure of He = 1200 kPa  x .20  = 240 kPa

Partial pressure of CH₄ = 1200 kPa  x  .6043  = 725.16 kPa

Partial pressure of C₂H₆ = 1200 kPa  x .108    = 129.6 kPa

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