Titanium reacts with iodine to form titanium(III) iodide, emitting heat.
2 Ti(s) + 3 I2(g) => 2 TiI3(s) ΔH°rxn= -839 kJ
Determine the masses of titanium and iodine that react if 1.55 X 10^3 kJ of heat is emitted by the reaction.
The enthalpy of rxn given (-839 KJ) is for two mols of Ti and three mols of I2.
First we determine the proportion of rxn that yields that would yield -1.55E3 kJ. Divide 1.55E3 by 839...
1.55E3 / 839 = 1.8474374
We multiply both the 2 mol Ti and 3 mol I2 by this factor to find the mols of reactants that were used...
2 mol Ti * 1.8474374 = 3.69487485
3 mol I2 * 1.8474374 = 5.54231227
Next convert mols to mass for each...
molar mass Ti = 47.867 g/mol.........3.69487485 * 47.867 = 177 g Ti (wth sigfigs in mind)
molar mass I2 = 253.8 g/mol............5.54231227 * 253.8 = 1410 g I2 (wth sigfigs in mind)
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