## Chemistry 10th Edition

- 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 amount in moles: $$200.0 \space g \times \frac{1 \space mole}{ 310.18 \space g} = 0.6448 \space mole$$ - Calculate or find the molar mass for $H_2SO_4$: $H_2SO_4$ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 4 )+ ( 32.07 $\times$ 1 )= 98.09 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$133.5 \space g \times \frac{1 \space mole}{ 98.09 \space g} = 1.361 \space moles$$ Find the amount of product if each reactant is completely consumed. $$0.6448 \space mole \space Ca_3(PO_4)_2 \times \frac{ 2 \space moles \ CaSO_4 }{ 1 \space mole \space Ca_3(PO_4)_2 } = 1.290 \space moles \space CaSO_4$$ $$1.361 \space moles \space H_2SO_4 \times \frac{ 2 \space moles \ CaSO_4 }{ 2 \space moles \space H_2SO_4 } = 1.361 \space moles \space CaSO_4$$ Since the reaction of $Ca_3(PO_4)_2$ produces less $CaSO_4$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $CaSO_4$: $CaSO_4$ : ( 40.08 $\times$ 1 )+ ( 16.00 $\times$ 4 )+ ( 32.07 $\times$ 1 )= 136.15 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$1.290 \space mole \space CaSO_4 \times \frac{ 136.15 \space g}{1 \space mole} = 175.6 \space g \space CaSO_4$$ --------- $$0.6448 \space mole \space Ca_3(PO_4)_2 \times \frac{ 1 \space mole \ Ca(H_2PO_4)_2 }{ 1 \space mole \space Ca_3(PO_4)_2 } = 0.6448 \space mole \space Ca(H_2PO_4)_2$$ - Calculate or find the molar mass for $Ca(H_2PO_4)_2$: $Ca(H_2PO_4)_2$ : ( 40.08 $\times$ 1 )+ ( 1.008 $\times$ 4 )+ ( 16.00 $\times$ 8 )+ ( 30.97 $\times$ 2 )= 234.05 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.6448 \space mole \space Ca(H_2PO_4)_2 \times \frac{ 234.05 \space g}{1 \space mole} = 150.9 \space g \space Ca(H_2PO_4)_2$$ - The superphosphate mass is equal to the sum of both products' masses: $$175.6 \space g + 150.9 \space g = 326.5 \space g$$