What’s the complete thermochemical equation for this reaction : I have 1.95g Na2CO3, 125.0mL of 2.50 mol/L of HCl.Initial temperature is 20.7 degrees and the final temperature is 22.5 degrees. The balanced equation is 2HCl+Na2CO3 arrow 2NaCl+CO2+H2O
Hint the answer is 52KJ
First, calculate the moles of Na2CO3 used:
1.95g Na2CO3 x (1 mol Na2CO3 / 105.99g Na2CO3) = 0.0184 mol Na2CO3
Next, calculate the moles of HCl used:
125.0mL HCl x 2.50 mol/L = 0.3125 mol HCl
Since the balanced equation shows a 1:1 stoichiometric ratio between Na2CO3 and HCl, the limiting reactant is Na2CO3. Therefore, 0.0184 mol Na2CO3 completely reacts with 0.0184 mol HCl.
Calculate the enthalpy change using the equation:
q = mcΔT = (125.0mL x 2.50 mol/L x 1.00g/mL x 4.18 J/g°C) x (22.5°C - 20.7°C) = 2620 J
Convert J to kJ:
2620 J = 2.62 kJ
Since 0.0184 mol Na2CO3 was used, the enthalpy change per mole is:
ΔH = 2.62 kJ / 0.0184 mol = 52 kJ
Therefore, the complete thermochemical equation for the reaction is:
2HCl + Na2CO3 → 2NaCl + CO2 + H2O ΔH = -52 kJ