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

35 g of Oxygen will give 59.7 g of SeO3

Explanation:

Atomic number of SeO3 = 34 + 3×16 =  82 g

In this reaction 3 moles of oxygen gives 2 moles of SeO3

therefore one mole of oxygen will give moles of SeO3

that is 32 g of oxygen will give g of SeO3

therefore 35 g of oxygen will give g of SeO3

=59.7 g of SeO3