Quesliui 2
jpl
consider the following chemical equation. how many moles of seo3 can be produced
from 35 g of oxygen and excess selenium?
25e + 302 - 25e03
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0.59 moles

Explanation:

Data Given

oxygen = 35 g

Moles of SeO₃ = ?

Reaction Given

          2Se + 3O₂ > 2SeO₃

Solution:

Step 1.

First to find grams of SeO₃

So, we know from reaction that 2 mole of selenium combine with 3 mole of oxygen and produces 2 mole of SeO₃

                         2Se  + 3O₂      > 2SeO₃

                        2mol    3 mol             2 mol

If we represent mole in grams

Then,

Molar mass of Se = 79 g/mol

Molar mass of O = 16 g/mol

Molar mass of SeO₃ = 79 + 3(16)

Molar mass of  SeO₃ = 127 g/mol

                        2Se              +     3O₂                  >       2SeO₃

                2mol (79 g/mol)     3 mol (32 g/mol)             2 mol (127 g/mol)

                        2Se   +  3O₂      > 2SeO₃

                        158 g    96 g             206 g

It is obvious from the reaction that 96 g of oxygen gives 206 g of SeO₃.

Now how many grams of SeO₃ will produce if 35 grams of oxygen react with excess of Selenium

Apply unity formula

                         96 g of O₂ ≅ 206 g of SeO₃

                           35 g of O₂ ≅ x g of SeO₃

By doing cross multiplication

                             g of SeO₃ = 206 x 35 / 96

                             g of SeO₃ =  75 g

Step 2.

Convert grams of SeO₃ to mole

Formula used

                    no. of moles = mass in grams (SeO₃) / Molar mass of SeO₃

Put values in above formula

                    no. of moles = 75 g / 127 g/mol

                    no. of moles = 0.59 mol

So,

35 g of oxygen produces 0.59 moles of SeO₃

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