Here's the question you clicked on:
yaaaimconfused15
How many grams of calcium phosphate can be produced when 89.3 grams of calcium chloride reacts with excess sodium phosphate? Unbalanced equation: CaCl2 + Na3PO4 → NaCl + Ca3(PO4)2 Show, or explain, all of your work along with the final answer.
Balanced equation: \[3CaCl_{2}+2Na_{3}PO_{4} \rightarrow 6NaCl+Ca_3(PO_4)_2\] Ca = 40.08g Cl = 35.45g Na = 11.99g P = 30.97g O = 16.00g It takes 3 moles of Calcium Chloride to produce 1 mole of Calcium Phosphate. \[\frac {1 \space mole \space CaCl_2}{(40.08 g \space Ca+(2\times35.45g \space Cl_2))} \times \frac{89.3g\space CaCl_2}1=0.805 \space moles\] \[\frac {3 \space moles \space CaCl_2}{1 \space mole \space Ca_3(PO_4)_2}=\frac{0.805 \space moles \space CaCl_2}{x \space moles \space Ca_3(PO_4)_2}\] \[3.00x =0.805\times1 \rightarrow x=\frac{0.805}{3.00} \rightarrow x=0.268\space moles \space Ca_3(PO_4)_2 \] \[89.3g \space of \space CaCl_2 \space will \space produce \space 72.45g \space of \space Ca_3(PO_4)_2\]
This is such a nice answer, I wish I could give it many medals more. Espex, good job on that. Yaaimconfused, please do give them a medal. This is another limiting reagent problem. There are others on this group so try to work them out as well. None of the solutions are as elegantly displayed as this one!!!