The heat of reaction has the wrong units...it should be -166.5kJ/MoleNa3PO4
So, the question is, the reaction calls for three moles (120g) of NaOH, so you will get 4/120 fraction of the heat of reaction.
Given:
H3PO4(aq) + 3 NaOH(aq) �¨ 3 H2O(l) + Na3PO4(aq)
ĢH = -166.5 kJ
What is the value for q (heat) if 4.00 g of NaOH reacts with an excess of H3PO4?
Please can you explain to me HOW to do this, not just give me an answer?
Thanks!
2 answers
Given:
H3PO4(aq) + 3 NaOH(aq) yields 3 H2O(l) + Na3PO4(aq)
change in H = -166.5 kJ
What is the value for q (heat) if 4.00 g of NaOH reacts with an excess of H3PO4?
Please can you explain to me HOW to do this, not just give me an answer?
Thanks!
H3PO4(aq) + 3 NaOH(aq) yields 3 H2O(l) + Na3PO4(aq)
change in H = -166.5 kJ
What is the value for q (heat) if 4.00 g of NaOH reacts with an excess of H3PO4?
Please can you explain to me HOW to do this, not just give me an answer?
Thanks!
2 answers