determine the heat released from the solution mass.
heat=(.1+.1)*cwater*(39.7-35)
Now look a the reaction:
3Ca(OH)2+2H3PO4>>Ca3(PO4)2+6H2O
Now looking at the reactants, you started with .03moles Ca(OH)2, and .07 mles H3PO4. So you have excess Ca(OH)2. The phopsporic acid used all .07moles.
So as a result, you got .035 moles of calcium phosphate.
Hr (kJ/mol)=Heatabove/.03
A 100.0 mL sample of 0.300 M Ca(OH)2 is mixed with 100.0 mL of 0.700 M H3PO4 in a coffee cup calorimeter.If both solutions were initially at 35.0 °C and the resulting solution had a temperature that was recorded at 39. 7°C, determine the ΔHrxn in units of kJ/mol. Assume the solution has the same specific heat and density as water.
2 answers
Ooops, that is .035 in the denominator