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You have an unknown that is a mixture of NaCl and Na3PO4 solids. You add 10.0 grams of the unknown solid to 190.0 grams of 2.000 M HCl solution with aninitial temperature of 18.92o C in a coffee cup calorimeter. The following reaction occurs: Na3PO4(s) + 3 HCl(aq)  3 NaCl(aq) + H3PO4(aq) ∆H = -499 kJ After the reaction is complete, you find the temperature in the cup to be 37.13o C. Note: addition of the NaCl solid to the HCl solution would have no heat released or absorbed; any temperature change is due solely to the reaction with Na3PO4 solid (the limiting reagent in the reaction above). What percent of the solid mixture is Na3PO4? You may make the following assumptions. The coffee cup acts as an isolated system. The resulting solution in the cup has a specific heat of 4.18 J/go C. The density of the resulting solution is 1.04 g/mL. Mol mass: Na3PO4 = 163.91g/mol, NaCl = 58.43g/mol, HCl = 36.46g/mol, H3PO4 = 97.99 g/mol