Answer:
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.
