In the synthesis of hydrogen iodide (HI), if 5 moles of hydrogen (H2) react with an excess of iodine (I2), calculate the number of moles of HI formed and the number of moles of iodine that remain unreacted.
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The balanced equation for the reaction between hydrogen and iodine to form hydrogen iodide is:
H2 + I2 -> 2HI
According to the stoichiometry of this equation, for every 1 mole of hydrogen (H2) reacting, 1 mole of iodine (I2) reacts to form 2 moles of hydrogen iodide (HI).
Therefore, if 5 moles of hydrogen react, the number of moles of hydrogen iodide formed is:
5 moles H2 x (2 moles HI / 1 mole H2) = 10 moles HI
So, 10 moles of hydrogen iodide is formed.
Since there is an excess of iodine in the reaction, all of the 5 moles of hydrogen will react, leaving no hydrogen unreacted. The number of moles of iodine that remain unreacted can be calculated by subtracting the moles of iodine that reacted from the initial moles of iodine:
Initial moles of iodine (I2) = 5 moles (since it is in excess)
Moles of iodine that reacted = 5 moles (since 5 moles of hydrogen reacted to form 5 moles of hydrogen iodide, which consumed 5 moles of iodine)
Moles of iodine that remain unreacted = Initial moles of iodine - Moles of iodine that reacted
= 5 moles - 5 moles
= 0 moles
So, none of the iodine remains unreacted.
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