Question

Consider the following chemical equation. How many moles of SeO3 can be produced from 35

g of oxygen and excess selenium?
25e +302 - 25e03

Answer 1

35 g of Oxygen will give 59.7 g of SeO3

Step-by-step 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
`(2)/(3)` moles of SeO3

that is 32 g of oxygen will give
`(2)/(3)* 82` g of SeO3

therefore 35 g of oxygen will give
`(2* 82* 35)/(3* 32)` g of SeO3

=59.7 g of SeO3