Respuesta :
Heat produced is given by -0.813 kJ
Calculation of heat produced:
The reaction:
[tex]3NaOH + H_3PO_4 \rightarrow Na_3PO_4 + H_2O[/tex] ΔH[tex]_{rxn}[/tex]= -173.7 kJ (Assumed)
Step 1: Calculation of moles using Molarity.
Volume of NaOH= 31.2 mL
Concentration= 0.45M
[tex]M=\frac{n}{V}\times1000\\\\n=\frac{M\times V}{1000} \\\\n= \frac{31.2\times0.45}{1000}\\n= 0.01404[/tex]of NaOH
Volume of [tex]H_3PO_4[/tex] = 65.4 mL
Concentration= 0.088M
[tex]n=\frac{0.088\times 65.4}{1000}\\\\n= 0.005755[/tex] mole of [tex]H_3PO_4[/tex]
Step 2: Limiting reagent
According to the above reaction
1 mole of [tex]H_3PO_4[/tex] reacts with 3 mole of NaOH
0.005755 moles will react with [tex]3\times 0.005775 \, NaOH[/tex]= 0.017 moles of NaOH is required. But we have 0.01404 moles of NaOH.
So NaOH is the limiting reagent.
Step3: Calculate ΔH:
We will limiting reagent to calculate the heat produced-
[tex]0.01404 \,mole \times\frac{-173.7}{3}= -0.813\, kJ[/tex]
Heat produced= -0.813 kJ
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