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Bond [772]
3 years ago
14

Suppose hydrochloric acid reacts with potassium sulfite yielding water, sulfur dioxide, and potassium chloride. Suppose 4 moles

of hydrochloric acid react with excess potassium sulfite. How many grams of sulfur dioxide are produced?
Chemistry
1 answer:
anzhelika [568]3 years ago
4 0

Answer:

128 grams of sulfur dioxide are produced.

Explanation:

2HCl+K_2SO_3\rightarrow SO_2+2KCl+H_2O

Moles of HCl = 4 moles

According to reaction, 2 moles of HCl gives 1 mole of sulfur dioxide gas.

Then 4 moles of HCl will give:

\frac{1}{2}\times 4 moles=2 moles of sulfur dioxide gas.

Mass of sulfur dioxide gas = 2 mol × 64 g/mol = 128 g

128 grams of sulfur dioxide are produced.

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If 36 000 kg of full cream milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream
Yuri [45]

Answer:

B=5522.33kg/h

C=478.11kg/h

Explanation:

Hi! It's a mass balance. First we have to determine the inflow.

mass flow rate = 36000kg / 6h = 6000kg / h

We define the input variable

- input flow (A) = 6000kg / h

-XgA = percentage of fat in A = 0.04

We define output variables.

- skim milk (B)

-creme (C)

-XgB = fat percentage at B = 0.0045

-XgC = percentage of fat in C = 0.45

Then we can start with the balance.

As a general rule, the mass balance is:

Input = Output

Balance sheet

1) A = B + C

Fat balance

2) A * XgA = B * XgB + C * XgC

Now we can solve.

We replace and clear B in equation 2

6000kg / h * 0.04 = B * 0.0045 + C * 0.45

B = (240kg / h) /0.045-C*0.45/0.0045

3) B = 53333.33kg / h-C * 100

We replace equation 3 in 1 and clear C

A = B + C

6000kg / h = 53333.33kg / h-C * 100 + C

C=(6000kg/h-53333.33kg/h)/(-99)

C=478.11kg/h

We replace C in equation 3 and calculate B

B = 53333.33kg / h-478.11kg/h * 100

B=5522.33kg/h

Then we have the values ​​of the outflows.

C=478.11kg/h

B=5522.33kg/h

7 0
3 years ago
Why do teapots have highly polished surface
Natali [406]

Good conductors of electricity and heat, form cations by loss of electrons,

6 0
3 years ago
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What is the molarity of a solution containing 60. grams of NaOH dissolved in 400. mL of water?
Hunter-Best [27]

Answer: 3.75 M

Explanation:

400 mL = 0.4 L

NaOH has a molar mass of around 40 g/mol.

\frac{60 grams}{40 g/mol} = 1.5 moles

Molarity = \frac{1.5 moles}{0.4 Liters} = 3.75 M

8 0
3 years ago
A student places three ice cubes in a beaker and allows them to partially melt. if she measures the temperature of the water in
atroni [7]

She will most likely observe that the temperature does not change during melting because the heat absorbed is used to overcome intermolecular forces rather than to increase the kinetic energy of the particles if she measures the temperature of the water in the beaker.

8 0
3 years ago
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Suppose you were preparing 1.0 L of a bleaching solution in a volumetric flask, and it calls for 0.21 mol of NaOCl. If all you h
yulyashka [42]

Answer:

0.256 L  

Explanation:

We should use the following formula:

concentration (1) × volume (1) =  concentration (2) × volume (2)

concentration (1) = 0.82 M NaOCl

volume (1) = ?

concentration (2) = 0.21 M NaOCl

volume (2) = 1 L

volume (1) = [concentration (2) × volume (2)] / concentration (1)

volume (1) = [0.21 / 1] / 0.82 = 0.256 L

3 0
3 years ago
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