Calculate the amount (mol) of the salt formed in the reaction: 90 g of lithium (I) oxide with orthophosphoric acid.

Given:
m (Li2O) = 90 g
To find:
n (Li3PO4) =?
Decision:
1) Let’s compose the equation of the chemical reaction, let the mass of the salt be x g.
3Li2O + 2H3PO4 = 2Li3PO4 + 3H2O
2) Find the molecular weights of lithium oxide and lithium phosphate.
M (Li2O) = 14 + 16 = 30 g / mol, since 3 lithium oxides are 90 g / mol
M (Li3PO4) = 21 + 31 + 64 = 116 g / mol, since salt 2, then 232 g / mol
3) 90 g: 90 g / mol = x g: 232 g / mol
x = (90 x 232): 90 = 232 g.
4) Find the amount of lithium phosphate substance.
n = m / M
n (Li3PO4) = 232 g: 116 g / mol = 2 mol.
Answer: n (Li3PO4) = 2 mol.