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3 years ago

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35g of HF was prepared by reacting 112g of CaF2 with an excess of H2SO4 calculate the percentage yield

1 answer:

Tom [10]3 years ago

7 0

Answer: 61%

The reaction equation should be

CaF2 + H2SO4 → 2HF + CaSO4

For every 1 molecule CaF2 used, there will be 2 molecules of HF formed. The molecular mass of CaF2 is 78/mol while the molecular mass of HF is 20g/mol. If the yield is 100%, the amount of HF formed by 112g CaF2 would be: 112g/(78g/mol) * 2 * (20g/mol)=57.43g

The percentage yield of the reaction would be: 35g/57.43g= 60.94%

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Kruka [31]

Answer:

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Explanation:

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The ICE table is then shown as:

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Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

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$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

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$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

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Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

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