Answer
326.5 g of superphosphate can be formed.
Work Step by Step
- 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:
$$ 1.290 \space mole \space CaSO_4 \times \frac{ 136.15 \space g}{1 \space mole} = 175.6 \space g \space CaSO_4$$
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$$ 0.6448 \space mole \space Ca_3(PO_4)_2 \times \frac{ 1 \space mole \ Ca(H_2PO_4)_2 }{ 1 \space mole \space Ca_3(PO_4)_2 } = 0.6448 \space mole \space Ca(H_2PO_4)_2 $$
- 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:
$$175.6 \space g + 150.9 \space g = 326.5 \space g$$
