Suppose the reaction Ca3(PO4)2 + 3H2SO4 --> 3CaSO4 + 2H3PO4 is carried out starting with
153 g of Ca3(PO4)2
and 76.8 g of H2SO4
How many grams of phosphoric acid will be produced.

The answer to your question is 51.2 g of H₃PO₄

Explanation:

Data

mass of Ca₃(PO₄)₂ = 153 g

mass of H₂SO₄ = 76.8 g

mass of H₃PO₄ = ?

Balanced chemical reaction

               Ca₃(PO₄)₂  +  3H₂SO₄  ⇒   3CaSO₄  +  2H₃PO₄

Process

1.- Calculate the molar mass of the reactants

Ca₃(PO₄)₂ = (3 x 40) + (2 x 31) + (8 x 16) = 120 + 62 + 128 = 310 g

H₂SO₄ = (1 x 2) + (32 x 1) + (16 x 4) = 2 + 32 + 64 = 98 g

2.- Calculate the limiting reactant

Theoretical yield Ca₃(PO₄)₂ / H₂SO₄ = 310 / 3(98) = 1.05

Experimental yield Ca₃(PO₄)₂ / H₂SO₄ = 153 / 76.8 = 1.99

The limiting reactant is H₂SO₄ because the experimental proportion was higher than the theoretical proportion.

3.- Calculate the molar mass of H₃PO₄

H₃PO₄ = (1 x 3) + (31 x 1) + (16 x 4) = 3 + 31 + 64 = 98 g

4.- Calculate the mass of H₃PO₄

                3(98) g of H₂SO₄ 2(98) g of H₃PO₄

                 76.8 g of H₂SO₄  x

                        x = (76.8 x 2 x 98) / (3 x 98)

                        x = 15052.8 / 294

                       x = 51.2 g of H₃PO₄

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.