Solution.
"m(HBr)=20.1g;"
"M(HBr)=81g\/mol;"
"\\nu(HBr)=\\dfrac{m(HBr)}{M(HBr)};"
"\\nu(HBr)=\\dfrac{20.1g}{81g\/mol}=0.248mol;"
The enthalpy of formation of hydrogen bromide is equal to -34.1kJ/mol, that is, in the formation of one mole of hydrogen bromide, 34 kJ of heat is absorbed.
"1mol----34kJ;"
"0.248mol--xkJ;"
"x=\\dfrac{0.248mol\\sdot34kJ}{1mol}=8.437kJ;"
Answer: 8.437kJ of heat is absorbed.
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