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lozanna [386]
3 years ago
11

A chemist prepares hydrogen fluoride by means of the following reaction:

Chemistry
1 answer:
natita [175]3 years ago
4 0

Answer:

39.3%

Explanation:

CaF2 + H2SO4 --> CaSO4 + 2HF

We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:

For CaF2;

Number of moles reacted= mass/molar mass

Molar mass of CaF2= 78.07 g/mol

Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride

Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride

0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride

Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass

Molar mass of hydrogen fluoride= 20.01 g/mol

Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)

Actual yield of HF was given in the question as 2.2g

% yield of HF= actual yield/ theoretical yield ×100

%yield of HF= 2.2/5.6 ×100

% yield of HF= 39.3%

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Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

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Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

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2) The density of the packed bed can be expressed as

\rho=f_v*\rho_v+f_s*\rho_s

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum  of the fractions ois equal to the total space).

Then we can rearrange

\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37

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