Answer:
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
