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Alex
2 years ago
11

THE SPENT ACID FROM A NITRATION PROCESS CONTAINS 33% H2SO4, 36% HNO3 AND 31% H2O. THIS IS TO BE STRENGTHENED BY THE ADDITION OF

CONCENTRATED H2SO4 CONTAINING 95% H2SO4 AND NITRIC ACID CONTAINING 78% HNO3. THE FINAL STRENGTHS OF MIXED ACID SOLUTION CONTAINS 40% H2SO4 43% HNO3 AND THE REST WATER. CALCULATE : A. THE AMOUNT OF SPENT ACID
B. THE AMOUNT OF CONCENTRATE H2SO4

C. THR AMOUNT OF CONCENTRATE HNO3 TO BE MIXED TO PRODUCE 1500 KG OF TGE DESIRED ACID

Chemistry
1 answer:
Nady [450]2 years ago
5 0

Answer:

a) 217.15 kg of spent acid

b) 556.14 kg of concentrate H₂SO₄

c) 726.70 kg of concentrate HNO₃

Explanation:

First, let's do a sketch for the problem (image below) and let's choose a basis of 1500 kg of production of the desired acid. Calling A the flow of spent acid, B the amount of concentrated H₂SO₄, C the flow of concentrated HNO₃, and P the flow of the product.

a) The global mass balance is:

input = output

A + B + C = P

A + B + C = 1500

The mass balance for H₂SO₄:

0.33A + 0.95B = 0.4x1500

0.33A + 0.95B = 600

0.95B = 600 - 0.33A

B = (600 - 0.33A)/0.95

The mass balance for HNO₃:

0.36A + 0.78C = 0.43x1500

0.36A + 0.78C = 645

0.78C = 645 - 0.36A

C = (645 - 0.36A)/0.78

Replacing B and C in the global mass balance:

A + (600 - 0.33A)/0.95 + (645 - 0.36A)/0.78 = 1500

0.1911A + 1458.5 = 1500

0.1911A = 41.5

A = 217.15 kg of spent acid

b) B = (600 - 0.33A)/0.95

B = (600 - 0.33x217.15)/0.95

B = 556.14 kg of concentrate H₂SO₄

c) C = (645 - 0.36A)/0.78

C = (645 - 0.36x217.15)/0.78

C = 726.70 kg of concentrate HNO₃

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Answer:

2

Explanation:

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2 years ago
Determine the poh of a 0.348 m ba(oh)2 solution at 25°c.
vladimir1956 [14]

<u>Given:</u>

Concentration of Ba(OH)2 = 0.348 M

<u>To determine:</u>

pOH of the above solution

<u>Explanation:</u>

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2 years ago
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expeople1 [14]

Hey there!:

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2 years ago
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2 years ago
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