Using the reaction below determine the amount of Sulfur proceduced in grams (this is a limiting reaction question) 2 H2S + SO2 -> 3 S + H2O 3.26g of dihydrogen sulfide react with 4.42 g of sulfur dioxide

= 2.2 g pf S. produced

Explanation:

Balanced Reaction equation:

→  

1 mole of H2S - 34.1g

? moles - 3.2g

= 3.2/34.1 = 0.09 moles of H2S

Also,

1 mole of S02 - 64.07 g

? moles - 4.42g

= 4.42/64.07 = 0.069 moles of SO2

Meaning SO2 is the limiting reagent

Finally, 3 moles of S -  32g of sulphur

0.069 mole = ? g of Sulphur

= 0.069 x 32

= 2.2 g pf S.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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