Answer to Question #86403 in General Chemistry for mike
1. Compute the mole fraction of H2 in the mixture.
2. The mixture is then allowed to stand, and the H2 starts to react with the I2 to give HI:
H2(g) + I2(g)----- 2HI(g)
At a certain point in the reaction, 6.60 mol of HI is present. Determine the mole fraction of H2 in the new mixture.
1. Mole fraction of H2 = n(H2)/n(mixture) = 10.2/(10.2+8.34)= 0.55 or 55%
2. H2 + I2 -> 2HI
If 6.60 mol of HI was formed, then amount of hydrogen requied was n(H2)= n(HI)/2= 6.60/2= 3.30 mol. Amount of iodine required is n(I2)= n(HI)/2= 6.60/2= 3.30 mol.
Find amout of a mixture of gases after reaction:
n(H2)= 10.2-3.30= 6.9 mol
n(I2)= 8.34-3.3= 5.04 mol
n(HI)= 6.60 mol
n(mixture)= 6.9+5.04+6.60= 18.54 mol
Mole fractionnof H2 = 6.9/18.54= 0.372 or 37.2%
Answer: 1. 55%, 2. 37.2%
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