If 15.0 mL of a 1.5M HCl solution at 22.5 degrees C is mixed with 25.0mL of a 1.5M NaOH solution at 21.5 degrees C that is in a calorimeter, and the final mixed solution temperature ends up at 28.5 degrees C,
1.)what is the balanced equation for this reaction?
2.) what is the source of the heat that is causing the increase in temperature?
3.) Calculate the amount of heat absorbed or lost for the HCl/NaOH solution. Assume aqueous conditions.(Cwater=4.18J/gxK, d=1.0g/mL)
1.
HCl + NaOH ==> NaCl + H2O
2.
H^+ + OH^= ==> H2O
3.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) assuming the calorimeter can be ignored. '
4. I'm surprised there is no 4th question to calculate kJ/mol for the heat of neutralization.
Sorry there is,
4.) Calculate the amount of het absorbed or lost by the calorimeter. The calorimeter's initial temperature is the same as the solution that is initially inside it.
5.) Determine the amount of heat absorbed or lost during this reaction.
6.)Which of the reactants is the limiting reagent? Determine the moles.
7.)Determine the amount of heat given off per mole of LR. Make sure to include an appropriate sign indicating whether it is an exothermic or endothermic process. Answer in kJ/mol
4. You don't have enough information to calculate the amount of heat absorbed/lost by the calorimeter. You need the calorimeter constant or data to calculate and you have neither.
5,6,7 depend upon 4.
1.) To determine the balanced equation for the reaction between HCl and NaOH, we need to identify the products and their stoichiometric ratios. In this case, HCl is a strong acid, and NaOH is a strong base. When they react, the products will be water (H2O) and sodium chloride (NaCl). The balanced equation is:
HCl + NaOH -> H2O + NaCl
2.) The source of the heat that is causing the increase in temperature is the exothermic nature of the reaction between HCl and NaOH. When acids and bases react, they undergo a neutralization reaction, and energy is released in the form of heat. The heat generated by this reaction is transferred to the surroundings, resulting in an increase in temperature.
3.) To calculate the amount of heat absorbed or lost for the HCl/NaOH solution, we can use the equation:
q = m * c * ΔT
where q is the heat (in joules), m is the mass (in grams), c is the specific heat capacity (in J/g·°C), and ΔT is the change in temperature (in °C).
First, we need to find the total mass of the solution. Since we are given the volumes and the density of water, we can calculate the mass using the formula:
mass = volume * density
mass(HCl solution) = 15.0 mL * 1.0 g/mL = 15.0 g
mass(NaOH solution) = 25.0 mL * 1.0 g/mL = 25.0 g
Next, we can determine the total heat exchanged using the formula:
q = (m(HCl solution) + m(NaOH solution)) * c * ΔT
Here, c = 4.18 J/g·°C (specific heat capacity of water) and ΔT = 28.5 °C - (22.5 °C + 21.5 °C) = 28.5 °C - 44.0 °C = -15.5 °C (since the reaction led to a decrease in temperature).
Using the given values:
q = (15.0 g + 25.0 g) * 4.18 J/g·°C * (-15.5 °C)
Calculating further:
q = 40.0 g * 4.18 J/g·°C * (-15.5 °C)
q ≈ -2494 J
Therefore, the amount of heat absorbed or lost for the HCl/NaOH solution is approximately -2494 J (since it is negative, it means that heat was lost during the reaction).