Answer to Question #62087 in General Chemistry for Nicole

Question

Suppose the reaction Ca3(PO4)2 + 3H2SO4 ® 3CaSO4 + 2H3PO4 is carried out starting with 119.79 g of Ca3(PO4)2 and 66.07 g of H2SO4. How much phosphoric acid will be produced?

Answer 1

119.79 g of Ca3(PO4)2 is 119.79 g/310 g/mol = 0.386 mol and 66.07 g of H2SO4 is 66.07 g/98 g/mol = 0.674 mol of it. According to the equation 1 mol of Ca3(PO4)2 needs some 3 mols of H2SO4. But it can be easily seen that H2SO4 is not enough and following calculations will be proceeded with using of H2SO4 amount. From 0.674 mols of H2SO4 we can get 2/3*0.674 mol of H3PO4 or 0.449 mol. 0.449 mol of H3PO4 is 0.449 mol*98 g/mol = 44.00 g.