Question

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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Answer 1

0.59 moles

Step-by-step 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