6K + B2O3 → 3K2O + 2B

How many moles of Boron can be produced if 3650g of B2O3 reacted?

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃    →    3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of  B₂O₃ with B from balance chemical equation:

                 B₂O₃          :          B

                    1              :          2

                52.42         :        2×52.42 = 104.84

Thus from 3650 g of  B₂O₃  104.84 moles of boron will produced.

the maximum mass of nh3 is nh3 in grams