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# Answer to Question #86403 in General Chemistry for mike

Question #86403
A gas mixture contains 10.2 mol of H2 and 8.34 mol of I2.

1. Compute the mole fraction of H2 in the mixture.
________ XH2=

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.
_______ XH2=

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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