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ANSWER
There are 3.3 moles of Cl2 in 1.1 moles of SeCl6
EXPLANATION
Given that
The number of moles of SeCl6 is 1.1 moles
Firstly, write a balanced equation of the reaction
[tex]\text{ SeCl}_6\text{ }\rightarrow\text{ Se + 3Cl}_2[/tex]In the reaction above, 1 mole SeCl6 gives 3 moles Cl2
Let x be the number of moles of Cl2
[tex]\begin{gathered} \text{ 1 mole SeCl}_6\text{ }\rightarrow\text{ 3 moles Cl}_2 \\ \text{ 1.1 mole SeCl}_6\text{ }\rightarrow\text{ x moles Cl}_2 \\ \text{ cross multiply} \\ \text{ 1 mole SeCl}_6\times\text{ x moles Cl}_2\text{ = 1.1 mole SeCl}_6\text{ }\times\text{ 3 moles Cl}_2 \\ \text{ Isolate x } \\ \text{ x = }\frac{1.1moles\cancel{SeCl_6}\times\text{ 3 moles Cl}_2}{1mole\cancel{SeCl_6}} \\ \\ \text{ x = 1.1 }\times\text{ 3} \\ \text{ x = 3.3 moles} \end{gathered}[/tex]Therefore, there are 3.3 moles of Cl2 in 1.1 moles of SeCl6
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