# Determine the masses of titanium and iodine that react if 1.66xx10^3 "kJ" of heat is emitted by the reaction?

## Titanium reacts with iodine to form titanium(III) iodide, emitting heat. $2 {\text{Ti"(s)+3"I"_2(g) -> 2"TiI}}_{3} \left(s\right)$ $\Delta {H}_{\text{rxn"^@ = -"839 kJ}}$

Jun 29, 2017

Here's what I got.

#### Explanation:

The problem provides you with the thermochemical equation that describes the synthesis of titanium(III) iodide.

$2 \text{Ti"_ ((s)) + 3"I"_ (2(g)) -> 2"TiI"_ (3(s))" "DeltaH_ "rxn"^@ = - "839 kJ}$

The thermochemical equation tells you that when $2$ moles of titanium react with $3$ moles of iodine gas, $2$ moles of titanium(III) iodide are produced and $\text{839 kJ}$ of heat are given off.

In your case, you know that $1.66 \cdot {10}^{3}$ $\text{kJ}$ we given off, so use the known enthalpy change of reaction to determine how many moles of titanium and of iodine gas reacted in order to allow the reaction to give off this much heat.

1.66 * 10^3 color(red)(cancel(color(black)("kJ"))) * "2 moles Ti"/(839 color(red)(cancel(color(black)("kJ")))) = "3.957 moles Ti"

1.66 * 10^3 color(red)(cancel(color(black)("kJ"))) * "3 moles I"_2/(839 color(red)(cancel(color(black)("kJ")))) = "5.936 moles I"_2

Finally, to convert the number of moles to grams, use the molar masses of the two reactants.

$3.957 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Ti"))) * "47.867 g"/(1color(red)(cancel(color(black)("mole Ti")))) = color(darkgreen)(ul(color(black)("189 g}}}}$

$5.936 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles I"_2))) * "253.81 g"/(1color(red)(cancel(color(black)("mole I"_2)))) = color(darkgreen)(ul(color(black)("1510 g}}}}$

The answers are rounded to three sig figs, the number of sig figs you have for the amount of heat given off by the reaction.