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2 years ago

Determine the enthalpy for this reaction: Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)

1 answer:

6 0

Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 kJ/mol
Al(OH)3(s)= −1277.0 kJ/mol
 H2O(l)= −285.8 kJ/mol
AlCl3(s) =−705.6 kJ/mol

Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
reactants products

products- reactants:

(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 kJ per mole at 25°C

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In the initial solution we have:

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$\frac{sugar}{water} = \frac{13.333}{y} = \frac{1}{50} \\y = 666.667 g$

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