66.667 mL of 3.000 M H2SO4 (aq) solution was neutralized by the stoichiometric amount of 4.000 M Al(OH)3 solution in a coffee cup calorimeter. The initial temperature of the solutions was 22.3 C and after mixing the temperature raised to 24.7 C. If the heat capacity of the coffee cup is 1.10 J/g C, calculate the delta H reaction in J/mol Al2(SO4)3

The heat of reaction is -6.7 KJ/mol.

The equation of the reaction is;

3H2SO4(aq)  + 2Al(OH)3 > Al2(SO4)3(s) + 6H2O(l)

Number of moles of H2SO4 = 66.667/1000 L ×  3.000 M = 0.2 moles

Number of moles of Al(OH)3 =  66.667/1000 L × 4.000 M = 0.27 moles

If 3 moles of H2SO4 reacts with 2 moles of Al(OH)3

0.2 moles of H2SO4 reacts with 0.2 moles × 2 moles/3 moles  = 0.13 moles

We can see that there is more than enough Al(OH)3 hence H2SO4 is the limiting reactant.

Temperature change =  24.7°C - 22.3°C = 2.4°C

Heat capacity of the coffee cup = 1.10 J/g C

Total volume of solution = 66.667 mL + 66.667 mL = 133.334 mL

Mass of solution = 133.334 g

Density of solution = 1 g/ml

Specific heat capacity of solution = 4.18 J/g C

Total heat released =- (Heat absorbed by calorimeter + Heat absorbed by solution)

Total heat released = -[(1.10 J/g C × 2.4°C) + (133.334 g × 4.18 J/g C × 2.4°C)]

Total heat released = -1.34 KJ

Heat of reaction = -1.34 KJ/0.2 moles = -6.7 KJ/mol

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- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

< 3

the objects have the same charge is correct.

nathan lee