___Li + ___AlPO4 --> ___Al + ___Li3PO4

If 14.5 grams of Lithium are reacted, how many moles of Aluminum are produced?

From 14.5 g of Li 0.7 moles of Al are produced.

Explanation:

Given data:

Moles of Li = 14.5 g

Moles of Al produced = ?

Solution:

Chemical equation:

3Li + AlPO₄   →    Al + Li₃PO₄

Number of moles of Li:

Number of moles = mass/ molar mass

Number of moles = 14.5 g/6.94 g/mol

Number of moles = 2.1

Now we will compare the moles of Al and Li from balance chemical equation.

                       Li         :         Al

                        3        :          1

                      2.1        :        1/3 ×2.1 = 0.7 mol

So from 14.5 g of Li 0.7 moles of Al are produced.

2b2t

explanation:

2b2t