The aluminum cup inside your calorimeter weighs 41.55 g. you add 59.21 g of 1.0 m acetic acid solution and 50.03 g of 1.0 m sodium hydroxide solution to the calorimeter. both solutions have an initial temperature of 19.9 oc, and the final temperature after addition is 26.8 oc. what is the molar enthalpy of neutralization, in units of kj/mol? assume that: the calorimeter is completely insulated the heat capacity of the empty calorimeter is the heat capacity of the aluminum cup: 0.903 j g-1 oc-1. the density of the two solutions is the same as that of water: 1.00 g/ml. the heat capacity of the two solutions is the same as that of water: 4.184 j g-1 oc-1.

The molar enthalpy of neutralization = 68.211 kJ/mol

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Heat can be calculated using the formula:

A calorimeter is a device used to measure the specific heat of material

The heat is not wasted into the environment because the calorimeter is a closed system, so the reaction heat is the same as the heat absorbed by the solution and the calorimeter

mass of solution = 59.21 + 50.03 = 109.24 g

c solution = 4,184 J / g ° C

ΔT = 26.8 ° C - 19.9 ° C = 6.9 ° C

Q = m. c ΔT

Q = 109.24. 4,184. 6.9

Q solution = 3153.7151  J

mass = 41.55 g

c the aluminum cup: 0.903 J / g ° C

ΔT = 26.8 ° C - 19.9 ° C = 6.9 ° C

Q calorimeter = m. c ΔT

Q calorimeter = 41.55. 0903. 6.9

Q calorimeter = 258.8856  J

So that:

reaction q = q rxn = - (q solution + q calorimeter)

q rxn = - (3153.7151+ 258.8856)

Q rxn = - 3412,6007 J

ΔH neutralization = change in enthalpy in the formation of 1 mole H₂O

Reactions that occur:

CH₃COOH (aq) + NaOH (aq) → CH₃COONa (aq) + H₂O (l)

The number of moles of H₂O formed is based on limiting reactants, so we look for moles each of the reactants

CH₃COOH = 59.21 g: 1 g / ml × 1.0 M = 59.21 mlmol = 0.05921 moles

NaOH = 50.03: 1 g / ml × 1.0 M = 50.03 mlmol = 0.05003 moles

NaOH mole is smaller, so it becomes a limiting reactant

So mole H₂O = mole NaOH = 0.05003 moles

then:

the difference between temperature and heat

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Specific heat

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relationships among temperature, heat, and thermal energy.

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When heat is added to a substance

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as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2