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The combustion of 0.4196 g of a hydrocarbon containing only C and H releases 17.55 kj of heat the masses of the products are CO2 equals 1.219 g and H2O equals 0.290 g?
Jguzman23
whats is the empirical formula of the compound? if the approximate molar mass of the compound is 76g/mol, calculate its standard enthalpy of formation
ANSWER
1.419 g CO_2_ * 1 mol CO_2_ /44.01 g CO_2_ *1mol C/1 mol CO_2_ = 0.03224 mol C
0.290 H_2_O * 1mol H_2_O/18.02 g H_2_O * 2 mol H/1mol H_2_O=0.03219 MOL H
Ratio is 1:1 so empirical formula is CH=13g/mol
molar mass 76g/mol ->76/13=6
molecular formula C_6_H_6_
2C_6_H_6_+15O_2_----->12CO_2_ + 6H_2_O
17.55kj/0.4196 C_6_H_6_* 78.11G C_6_H_6_/1mol C_6_H_6_ * 2 mol C_6_H_6_= 653 kj/mol (releases)
Hrxn=12H_f_CO_2_ + 6H_f_ H_2_O - 2 H_f_ C_6_H_6_ - 15
-6534kj/mol=(120(-39305kj/mol)+6(-285.8kj/mol)-(2)(H_f_ C_^_H_^_)-(15) (0)
H_f_(C_6_H_6_)=49kj/mol
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