2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(l) ΔH = -118 kJ ?
Consider the following reaction.
2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(l) ΔH = -118 kJ
Calculate the heat when 104.4 mL of 0.500 M HCl is mixed with 300.0 mL of 0.580 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0°C and that the final mixture has a mass of 404.4 g and a specific heat capacity of 4.18 J/°C·g, calculate the final temperature of the mixture.
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First its best to change mL to Liters, which is just dividing by 1000. Then, you find the limiting reactant:
.1044L x .500M HCL = .0522 mol HCl x 1molBaCl2/2molHCL = .0261 mol BaCl2
.300L x .580 M Ba(OH)2 = .174 mol Ba(OH)2
This shows that the limiting reagent is going to be HCl. This is always the first step you should take in reaction calculations. So, the change in enthalpy is going to be -118kJ x mols BaCl2 produced or
-118 x .0261 = 3.0798 kJ
So now we have the energy released from the reaction. All you need to do now is calculate how much that energy will change the temperature of the water.
104.4 mL + 300.0 mL = 404.4 mL = 404.4 g H2O
1mL of water = 1 g H2O, and we need the mass of water to calculate the change in temperature
404.4g x 4.18 J/C*g x delta T = 3.0798 x 1000 J (here we have to convert kJ to J by multiplying by 1000)
deltaT= +1.822
25.0 + 1.822 = 26.822 there's your answer