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Suppose the reaction Ca3(PO4)2 + 3H2SO4 → 3CaSO4 + 2H3PO4 is carried out starting with 151...

Suppose the reaction Ca3(PO4)2 + 3H2SO4 → 3CaSO4 + 2H3PO4 is carried out starting with 151 g of Ca3(PO4)2 and 69.0 g of H2SO4. How much phosphoric acid will be produced? 95.4 g 46.0 g 220.0 g 103.4 g 68.9 g

Solutions

Expert Solution

Ca3(PO4)2 + 3H2SO4 → 3CaSO4 + 2H3PO4

Molecular weight of Ca3(PO4)2 =310 g

Molecular weight of H2SO4 = 98 g

No of moles of Ca3(PO4)2 = 151/310

                                        = 0.5 M

No of moles of H2SO4 = 69/98

                                 = 0.7 M

According to the equation 3 moles of H2SO4 gives 2 moles of H3PO4

0.7 moles give?

No of moles of H3PO4 = 0.7*2/3

                                   = 0.46 M

0.46 M = wieght/molecular weight of H3PO4

0.46 = w/98

w = 0.46/98

    = 0.0047 g

Weight of H3PO4 produced = 0.007 g

queen_honey_blossom answered 2 years ago
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