1 mol of hydrogen gas and 1 mol of iodine vapor were mixed and allowed to react. After t seconds, 0.8 mol of hydrogen remained.
The number of moles of hydrogen iodide formed at t seconds was:
A) 0.2
B) 0.4
C) 0.8
D) 1.6
The answer is B. Please explain how you worked this out.
Answer (3)
At the beginning there was just 1 mole of hydrogen and **1 mole **of iodin. The reaction goes: H 2 + I 2 ⇒ 2 H I One mole of hydrogen goes with one mole of iodin, creating 2 moles of product. If 0.8 of hydrogen remaind, it means that 1-0.8=0.2 reacted. So: 1 mole H2 react with 1 mole I2 creating 2 moles of products. So 0.2 of hydrogen reacted with 0.2 mole of iodin creating 0.4 moles. 1 m o l e H 2 − − − 2 m o l es H I 0.2 m o l e − − − x x = 0.4
0.2 moles of hydrogen have reacted, and since the reaction produces hydrogen iodide in a 2:1 mole ratio, this leads to the formation of 0.4 moles of hydrogen iodide. B) 0.4 is the correct option.
The reaction between hydrogen gas (H₂) and iodine vapor (I₂) to form hydrogen iodide (HI) follows the balanced equation:
H₂(g) + I₂(g) → 2HI(g)
When the reaction takes place, both reactants are consumed in a 1:1 mole ratio to produce hydrogen iodide at a 2:1 mole ratio. If after time t seconds, 0.8 moles of hydrogen remains, then 0.2 moles of hydrogen have reacted (since we started with 1 mole). According to the stoichiometry of the reaction, this would lead to the formation of twice as many moles of HI because of the 1:2 ratio indicated by the balanced equation. Thus, 0.4 moles of HI are produced.
After 0.2 moles of hydrogen reacted, according to the balanced equation, 0.4 moles of hydrogen iodide were formed, making option B the correct choice. The stoichiometry shows that for every mole of H₂ reacted, two moles of HI are produced. Therefore, with 0.2 moles of H₂ reacted, we have 0.4 moles of HI produced.
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