Find the mass of the precipitate released during the interaction of 20g. Lithium oxide with phosphoric acid.

| June 21, 2021 | education

To solve the problem, we compose the reaction equation:
3Li2O + 2H3PO4 = 2Li3PO4 + 3H2O – ion exchange reaction, precipitate of lithium phosphate is precipitated;
Let’s determine the molar masses of substances:
M (Li2O) = 6.9 * 2 + 16 = 29.8 g / mol;
M (Li3PO4) = 6.9 * 3 + 30.9 + 16 * 4 = 115.6 g / mol;
Let’s calculate the number of moles of lithium oxide by the formula:
Y (Li2O) = m / M = 20 / 29.8 = 0.67 mol;
Determine the number of moles of lithium phosphate salt:
0.67 mol (Li2O) – X mol (Li3PO4);
-3 mol -2 mol from here, X mol (Li3PO4) = 0.67 * 23 = 0.446 mol;
Let’s make a calculation using the formula, calculating the mass of lithium phosphate:
m (Li3PO4) = 0.446 * 115.6 = 51.55 g.
Answer: the mass of the precipitated lithium phosphate precipitate is 51.55 g.



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