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tonya1807 tonya1807
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12 years ago
How do I solve this? What steps do I take?

For the reaction: 3CaCl2 (aq) + 2Na3PO4 (aq) Rightwards Arrow Ca3(PO4)2 (s) + 6NaCl (aq)

Identify the limiting and excess reactants in a reaction mixture containing 50.0 g of calcium chloride and 100.0 g of sodium phosphate. Also, how many grams are left over of the excess reactant when the reaction is over?
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12 years ago
3CaCl2 (aq) + 2Na3PO4 (aq) Rightwards Arrow Ca3(PO4)2 (s) + 6NaCl (aq)

First think about the stoichiometry equation and calculate the number of moles of each reaction.

Number of moles of CaCl2 = 50 g / 110.98 gmol^-1 = 0.4505 mol
Number of moles of Na3PO4 = 100 g / 163.9408 gmol^-1 = 0.61 mol

The mole ratio of CaCl2 : Na3PO4 = 3:2
Therefore the number of moles of Na3PO4 to react with CaCl2 = (0.4505/3) * 2 = 0.300mol
There is enogh moles of phosphate to react.

The number of moles of CaCl2 to react with Na3PO4 = (0.61/2)* 3 = 0.915mol
There is only 0.4505 moles of calcium chloride. So, the reaction undergoes till the all calcium chloride reacts. There is no sufficient amount to react till all phosphates consume. Therefore the limiting reagent is calcium chloride.
Till limiting reagent is consumed completely reaction happens. But after no reaction.

The number of moles of Na3PO4 remain in the system = 0.61- 0.30 = 0.31 mol
The weightof excess reagent remain = 0.31mol * 163.9408gmol^-1 = 50.82g
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