Calcium phosphate weighing 62 g was treated with concentrated sulfuric acid.

Calcium phosphate weighing 62 g was treated with concentrated sulfuric acid. Calculate the mass of the phosphoric acid formed.

Given:
m (Ca3 (PO4) 2) = 62 g

Find:
m (H3PO4) -?

Solution:
1) Write the equation of a chemical reaction:
Ca3 (PO4) 2 + 3H2SO4 => 3CaSO4 + 2H3PO4;
2) Calculate the molar masses:
M (Ca3 (PO4) 2) = 310 g / mol;
M (H3PO4) = 98 g / mol;
3) Calculate the amount of calcium phosphate substance:
n (Ca3 (PO4) 2) = m (Ca3 (PO4) 2) / M (Ca3 (PO4) 2) = 62/310 = 0.2 mol;
4) Determine the amount of phosphoric acid substance:
n (H3PO4) = n (Ca3 (PO4) 2) * 2 = 0.2 * 2 = 0.4 mol;
5) Calculate the mass of phosphoric acid:
m (H3PO4) = n (H3PO4) * M (H3PO4) = 0.4 * 98 = 39.2 g.

Answer: The mass of H3PO4 is 39.2 g.




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