## Chemistry 10th Edition

The maximum amount of $Ca_3(PO_4)_2$ is equal to 18.0 g
- Calculate or find the molar mass for $Ca(OH)_2$: $Ca(OH)_2$ : ( 40.08 $\times$ 1 )+ ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 2 )= 74.10 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$12.9 \space g \times \frac{1 \space mole}{ 74.10 \space g} = 0.174 \space mole$$ - Calculate or find the molar mass for $H_3PO_4$: $H_3PO_4$ : ( 1.008 $\times$ 3 )+ ( 16.00 $\times$ 4 )+ ( 30.97 $\times$ 1 )= 97.99 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$18.37 \space g \times \frac{1 \space mole}{ 97.99 \space g} = 0.1875 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.174 \space mole \space Ca(OH)_2 \times \frac{ 1 \space mole \ Ca_3(PO_4)_2 }{ 3 \space moles \space Ca(OH)_2 } = 0.0580 \space mole \space Ca_3(PO_4)_2$$ $$0.1875 \space mole \space H_3PO_4 \times \frac{ 1 \space mole \ Ca_3(PO_4)_2 }{ 2 \space moles \space H_3PO_4 } = 0.09375 \space mole \space Ca_3(PO_4)_2$$ Since the reaction of $Ca(OH)_2$ produces less $Ca_3(PO_4)_2$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $Ca_3(PO_4)_2$: $Ca_3(PO_4)_2$ : ( 40.08 $\times$ 3 )+ ( 16.00 $\times$ 8 )+ ( 30.97 $\times$ 2 )= 310.18 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.0580 \space mole \times \frac{ 310.18 \space g}{1 \space mole} = 18.0 \space g$$