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

2b2t

explanation:

2b2t

answer: cellular respiration takes in food and uses it to create atp, a chemical which the cell uses for energy. usually, this process uses oxygen, and is called aerobic respiration. it has four stages known as glycolysis, link reaction, the krebs cycle, and the electron transport chain.