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# Separation Processes MCQ’s – Rate Based Model for Leaching

If the density of CuO in ore ρ = 0.05

It is leached with H2SO4 = 0.5M

Ore particle have radius r=0.5

Effective diffusivity D_{e}= 6*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 3.8h

b) 4.6h

c) 7.8h

d) 6.4h

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Answer: a

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=3.79h.

If the density of CuO in ore ρ = 0.05

It is leached with H2SO4 = 0.6M

Ore particle have radius r=0.5

Effective diffusivity D_{e}= 6*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 3.8h

b) 4.56h

c) 7.87h

d) 9.4h

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Answer: b

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=4.56h.

If the density of CuO in ore ρ = 0.05

It is leached with H2SO4 = 0.5M

Ore particle have radius r=0.56

Effective diffusivity D_{e}= 6*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 4.2h

b) 4.3h

c) 4.76h

d) 6.4h

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Answer: c

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=4.76h.

If the density of CuO in ore ρ = 0.08

It is leached with H2SO4 = 0.5M

Ore particle have radius r=0.5

Effective diffusivity D_{e}= 6*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 6.08h

b) 6.32h

c) 6.35h

d) 6.4h

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Answer: a

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=6.08h.

If the density of CuO in ore ρ = 0.06

It is leached with H2SO4 = 0.5M

Ore particle have radius r=0.7

Effective diffusivity D_{e}= 6*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 10.2h

b) 10.42h

c) 10.54h

d) 6.4h

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Answer: b

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=10.42h.

If the density of CuO in ore ρ = 0.03

It is leached with H2SO4 = 0.5M

Ore particle have radius r=0.5

Effective diffusivity D_{e}= 6*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 2.1h

b) 2.28h

c) 7.8h

d) 6.4h

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Answer: b

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=2.28h.

If the density of CuO in ore ρ = 0.04

It is leached with H2SO4 = 0.5M

Ore particle have radius r=0.5

Effective diffusivity D_{e}= 6*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 3.8h

b) 4.75h

c) 7.8h

d) 6.4h

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Answer: b

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=4.75h.

If the density of CuO in ore ρ = 0.05

It is leached with H2SO4 = 0.5M

Ore particle have radius r=0.66

Effective diffusivity D_{e}= 6*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 6.12h

b) 6.22h

c) 6.62h

d) 6.4h

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Answer: c

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=6.62h.

If the density of CuO in ore ρ = 0.05

It is leached with H2SO4 = 0.5M

Ore particle have radius r=0.5

Effective diffusivity D_{e}= 4.44*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 5.13h

b) 5.3h

c) 5.65h

d) 6.4h

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Answer: a

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=5.13h.

If the density of CuO in ore ρ = 0.05

It is leached with H2SO4 = 0.5M

Ore particle have radius r=0.5

Effective diffusivity D_{e}= 9*10-6

Molecular weight M_{b}=79.6

B=0.5

C_{ab}=0.01

Calculate the time required for total leaching.

a) 2.53h

b) 3.4h

c) 5.56h

d) 6.78h

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Answer: a

Explanation: T= ρr2/6D_{e}M_{b}bC_{ab}, hence t=2.53h.