Youhcl + naoh --> nacl + h2o. 200cm^3 of 0.2 mol/dm^3 hcl was reacted with 100 cm^3 of 0.15 mol/dm^3 naoh, where the enthalphy change is -57.3 kJ. find the overall enthalpy change using moles
Question
Youhcl + naoh --> nacl + h2o. 200cm^3 of 0.2 mol/dm^3 hcl was reacted with 100 cm^3 of 0.15 mol/dm^3 naoh, where the enthalphy change is -57.3 kJ. find the overall enthalpy change using moles
Solution 1
The first step is to determine the number of moles of each reactant.
For HCl, the number of moles is calculated by multiplying the volume (in dm^3) by the concentration (in mol/dm^3). So, 0.2 dm^3 * 0.2 mol/dm^3 = 0.04 mol.
For NaOH, the number of moles is calculated in the same way: 0.1 dm^3 * 0.15 mol/dm^3 = 0.015 mol.
The balanced chemical equation shows that one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of H2O. Therefore, the limiting reactant is NaOH because we have less moles of it.
The enthalpy change given (-57.3 kJ) is for the reaction as written, which we've determined is based on one mole of NaOH reacting.
Since we have 0.015 mol of NaOH, the overall enthalpy change for this reaction is -57.3 kJ/mol * 0.015 mol = -0.8595 kJ.
Solution 2
The first step is to determine the number of moles of each reactant.
For HCl, the number of moles is calculated by multiplying the volume (in dm^3) by the concentration (in mol/dm^3). So, 0.2 dm^3 * 0.2 mol/dm^3 = 0.04 mol of HCl.
For NaOH, the number of moles is calculated in the same way: 0.1 dm^3 * 0.15 mol/dm^3 = 0.015 mol of NaOH.
The balanced chemical equation shows that one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of H2O. Therefore, the limiting reactant is NaOH because we have less moles of it.
The enthalpy change given (-57.3 kJ) is for the reaction as written, which we've determined is based on one mole of NaOH reacting.
Since we have 0.015 mol of NaOH, the overall enthalpy change for this reaction is -57.3 kJ/mol * 0.015 mol = -0.8595 kJ.
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