Question

Consider the reaction 3CaCl2(aq) + 2Na3PO4(aq) --> Ca3(PO4)2(s) + 6 NaCl(aq) (molar mass of calcium phosphate is 310.18 g/mol)

What mass of Ca3(PO4)2 is produced from 94.2 mL of 0.250 M Na3PO4 solution and 275 mL of 0.150 M CaCl2 solution? _____ g (Give answer to correct sig figs in units of g)

Answer 1

Here the given reaction is ,

$3CaCl_2(aq) +2Na_3PO_4(aq)\longrightarrow Ca_3(PO_4)_2(s)+6 NaCl$

Here amount of $Na_3PO_4$ is 94.2 mL of 0.25M solution

moles of Na₃PO₄= $Molarity\times volume (in L)=0.0942\times0.25=0.02355$ mole

similarly , moles of $CaCl_2= molarity\times volume= 0.15\times 0.275= 0.04125$ mole

from above reaction we can say that, 3 mole of CaCl₂ react with 2 mole of Na₃PO₄

0.04125 mole of CaCl₂ react with = $\frac{2}{3} \times 0.04125=0.0275$ mole so from here we are

getting the limiting reagent as Na₃PO₄

so from the above reaction we can say that ,

2 mole of $Na_3PO_4$ gives 1 mole of $Ca_3(PO_4)_2$

so, 0.02355 mole of $Na_3PO_4$ gives = $\frac{0.02355}{2}= 0.011775$ mole of $Ca_3(PO_4)_2$

so mass of $Ca_3(PO_4)_2= mole\times molar mass= 0.011775\times 310= 3.6425$ g