2. In a calorimetry experiment, 50.0 g of a 2.04 mole/kg HCl solution is mixed with 50.0 g of a 2.13
mol/kg NaOH solution. The specific heat of the resulting solution is 3.90 J·˚C−1·g−1.
The temperature of the mixture rises from 22.1 ºC to 37.0 ºC. Calculate the enthalpy change for the
overall reaction
3^x=45/79000^y^b
that does not help
"helper" doesn't seem to get how unhelpful it is to spew out a bunch of numbers with no explanation.
Carmin, I suggest you repost your question, and simply put "Chemistry" in the School Subject box.
helper look for my question and help me please
To calculate the enthalpy change for the overall reaction, we can use the equation:
ΔH = (m × c × ΔT) / n
Where:
ΔH is the enthalpy change for the overall reaction
m is the mass of the solution (in grams)
c is the specific heat capacity of the solution (in J·˚C−1·g−1)
ΔT is the change in temperature (in ˚C)
n is the number of moles of the limiting reactant in the reaction
First, we need to calculate the number of moles of the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
To find the limiting reactant, we need to compare the number of moles of HCl and NaOH. The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H₂O
The stoichiometric ratio between HCl and NaOH is 1:1. This means that for every 1 mole of HCl, we need 1 mole of NaOH. Since the moles of HCl is given as 2.04 mole/kg and the moles of NaOH is given as 2.13 mol/kg, we can compare the number of moles of HCl and NaOH by multiplying the moles per kg by the mass of the solutions:
Moles of HCl = (2.04 mol/kg) × (0.05 kg) = 0.102 mol
Moles of NaOH = (2.13 mol/kg) × (0.05 kg) = 0.1065 mol
Since the stoichiometric ratio is 1:1, the limiting reactant is HCl because it has the smaller amount of moles.
Next, we can calculate the enthalpy change using the equation mentioned earlier:
ΔH = (m × c × ΔT) / n
Substituting the given values:
m = 50.0 g (mass of the solution)
c = 3.90 J·˚C−1·g−1 (specific heat of the solution)
ΔT = (37.0 ºC - 22.1 ºC) = 14.9 ºC (change in temperature)
n = 0.102 mol (number of moles of HCl, the limiting reactant)
ΔH = (50.0 g × 3.90 J·˚C−1·g−1 × 14.9 ºC) / 0.102 mol
Calculating this value will give you the enthalpy change for the overall reaction.