How much orthophosphoric acid can be obtained by the interaction of 620 g

How much orthophosphoric acid can be obtained by the interaction of 620 g of calcium orthophosphate and 196 g of sulfuric acid, if the yield of the reaction product is 80%?

We implement the solution:
1. According to the condition of the problem, we compose the equation:
Ca3 (PO4) 2 + 3H2SO4 = 2H3PO4 + 3CaSO4 – ion exchange reaction, phosphoric acid was obtained;
2. Let’s make the calculations:
M Ca3 (PO4) 2 = 310 g / mol;
M (H2SO4) = 98 g / mol;
M (H2SO4) = 98 g / mol.
3. Determine the number of moles of starting materials:
Y Ca3 (PO4) 2 = m / M = 620/310 = 2 mol;
Y (H2SO4) = m / M = 196/98 = 2 mol.
4. Proportion:
2 mol Ca3 (PO4) 2 – X mol (Н3PO4);
-1 mol – 2 mol from here, X mol (H3PO4) = 2 * 2/1 = 4 mol.
5. Find the mass of the product:
m (H3PO4) = Y * M = 4 * 98 = 392 g (theoretical weight).
W = m (practical) / m (theoretical) * 100;
m (H3PO4) = 0.80 * 392 = 313.6 g (practical weight).
Answer: orthophosphoric acid was obtained, the mass of which is 313.6 g.