Problem: Consider the following unbalanced equation:Ca3(PO4)2 (s) + H2SO4 (aq) → CaSO4 (s) + H3PO4 (aq)What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% H2SO4 by mass)?

To balance equation → number of atoms must be the same for both sides of reaction → add coefficients

Balanced: Ca₃(PO₄)_2(s) + 3H₂SO_4(aq) → 3CaSO_4(s) + 2H₃PO_4(aq)

Reactants                        Products

Ca - 3                                     Ca – 1 3

P - 2                                       P - 1 2

O - 12 20                                O - 8 20

H -2 6                                     H - 3 6

S - 1 3                                    S - 1 3

$$\begin{gathered} \boxed{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{p}\mathbf{e}\mathbf{r}\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{,}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{=}\mathbf{ }\frac{\mathbf{mass}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{ }\mathbf{ }}{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}\mathbf{\times }\mathbf{100}} \\ \mathbf{mass}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{kg}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{3}}\mathbf{g}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{kg}}}\mathbf{\times }\frac{\mathbf{98}}{\mathbf{100}} \end{gathered}$$

mass H₂SO₄  = 980 g H₂SO₄

Determine limiting reactant → reactant that forms the less amount of product. 

• determines the maximum mass of the product (H₂O) formed.

Mole to mole comparison: 

• 3 mol H₂SO₄ forms 3 mol CaSO₄
• 3 mol H₂SO₄ forms 2 mol H₃PO₄