Question

Using the unbalanced equation above calculate the moles of hydrogen iodide vapor produced by 2.8 moles of iodine vapor.

_H2(g) + _l2(g) > _Hl (g)

Answer 1

# Using unbalanced chemical equation :

The reaction is :

$H_2+I_2\Rightarrow HI$

According to the equation:

1 mole of iodide vapor gave 1 mole of Hydrogen iodide

Hence 2.8 mole of iodide vapor gave 2.8 moles of Hydrogen iodide.

So the answer is 2.8 mole of Hydrogen iodide,if the equation is unbalanced.

# Using balanced chemical equation :

If we balance the equation then it becomes

$H_2+I_2 \Rightarrow2HI$

Here in this case:

1 mole of iodide vapor gave 2 mole of Hydrogen iodide

Hence 2.8 mole of iodide vapor gave 2.8$\times2=5.6$ moles of Hydrogen iodide.

So the answer is 5.6 mole of Hydrogen iodide,if the equation is balanced.

Thank you...