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Studentka2010 [4]

3 years ago

7

At this temperature, 0.400 mol of h2 and 0.400 mol of i2 were placed in a 1.00-l container to react. what concentration of hi is

present at equilibrium?

1 answer:

Mnenie [13.5K]3 years ago

3 0

Kc = [HI]² / {[H2] [I2]}
53.3 = (2x)² / {(0.400M - x)(0.400M - x)}
sqrt(53.3) = 2x / (0.400M - x)
2.92 - 7.30x = 2x
x = 0.314M

[HI] = 2x = 0.628M

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<u>Answer: </u>

<u>For 1:</u> The volume of HCl required is 6 L.

<u>For 2:</u> The volume of HCl required is 9 L.

<u>For 3:</u> The volume of sulfuric acid required is 4.5 L.

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$ ......(1)

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base

<u>For 1:</u>

We are given:

$n_1=1\\M_1=1M\\V_1=?L\\n_2=1\\M_2=3.0M\\V_2=2.0L$

Putting values in equation 1, we get:

$1\times 1\times V_1=1\times 3\times 2\\\\V_1=6L$

Hence, the volume of HCl required is 6 L.

<u>For 2:</u>

We are given:

$n_1=1\\M_1=1M\\V_1=?L\\n_2=2\\M_2=3.0M\\V_2=1.5L$

Putting values in equation 1, we get:

$1\times 1\times V_1=2\times 3.0\times 1.5\\\\V_1=9L$

Hence, the volume of HCl required is 9 L.

<u>For 3:</u>

To calculate the volume of acid, we use the equation:

$N_1V_1=N_2V_2$

where,

$N_1\text{ and }V_1$ are the normality and volume of acid

$N_2\text{ and }V_2$ are the normality and volume of base

We are given:

$N_1=1.0N\\V_1=?L\\N_2=3.0N\\V_2=1.5L$

Putting values in above equation, we get:

$1.0\times V_1=3.0\times 1.5\\\\V_1=4.5L$

Hence, the volume of sulfuric acid required is 4.5 L.

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