Consider the following reaction.
2 HCl(aq) + Ba(OH)2(aq) ----> BaCl2(aq) + 2 H2O(l) ΔH = -118 kJ
Calculate the heat when 107.6 mL of 0.500 M HCl is mixed with 300.0 mL of 0.565 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 407.6 g and a specific heat capacity of 4.18 J/°C·g, calculate the final temperature of the mixture.
_________ °C
2 HCl(aq) + Ba(OH)2(aq) ----> BaCl2(aq) + 2 H2O(l) ΔH = -118 kJ
Calculate the heat when 107.6 mL of 0.500 M HCl is mixed with 300.0 mL of 0.565 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 407.6 g and a specific heat capacity of 4.18 J/°C·g, calculate the final temperature of the mixture.
_________ °C
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2HCl(aq) + Ba(OH)2(aq) ----> BaCl2(aq) + 2H2O(l) ΔH = -118 kJ
0.1076L x 0.5M = 0.0538moles HCl
0.3L x 0.565 Ba(OH)2 = 0.17moles Ba(OH)2
0.0538moles HCl requires 1/2 x Ba(OH)2 or 0.0269moles Ba(OH)2.....HCl = limiting reactant
0.0568moles HCl x (-118kJ / 2moles HCl) = -3.17kJ released with 0.0568moles HCl
Q = mass x spefic heat x deltaT
3174.2J = 407.6g x 4.184J/g-C x deltaT
deltaT = 1.86C
new temperature = 26.86C
0.1076L x 0.5M = 0.0538moles HCl
0.3L x 0.565 Ba(OH)2 = 0.17moles Ba(OH)2
0.0538moles HCl requires 1/2 x Ba(OH)2 or 0.0269moles Ba(OH)2.....HCl = limiting reactant
0.0568moles HCl x (-118kJ / 2moles HCl) = -3.17kJ released with 0.0568moles HCl
Q = mass x spefic heat x deltaT
3174.2J = 407.6g x 4.184J/g-C x deltaT
deltaT = 1.86C
new temperature = 26.86C