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3 years ago

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You start your experiment with an empty test tube having a mass of 10.3362 g. You are working with a hydrated compound of calciu

m sulfate. After adding the hydrate to the empty test tube, the mass of the test tube and hydrate is 14.5549 g. After heating the hydrate, you weigh the test tube again and the mass is 13.6737 g. Determine the number of moles of water in the original hydrated calcium sulfate compound. Hint: The hydrate will be calcium sulphate Xhydrate. You must determine the number for X.

1 answer:

ASHA 777 [7]3 years ago

3 0

Answer:

0.04896 moles of water

Explanation:

The mass of the test tube is 10.3362 g

The mass of the test tube + hydrated calcium sulfate is 14.5549 g

The mass of the test tube + calcium sulfate is 13.6737 g

Therefore substracting these you can get the mass of hydrated calcium sulfate which is:

14.5549 - 10.3362 = 4.2187 g

The mass of calcium sulfate:

13.6737 - 10.3362 = 3.3375 g

And finally the mass of water in the hydrated compound:

4.2187 - 3.3375 = 0.8812 g

The molar mass of water is 18 g/mol, then the number of moles is:

moles of water = 0.8812 / 18 = 0.04896 moles

Now to find the X in calcium sulphate Xhydrate

You have to find the proportion in moles between calcium sulphate and water.

The mass of calcium sulphate is 3.3375 g and the molar mass is 136.14 g/mol. With that you calculate the moles:

moles of calcium sulphate = 3.3375 / 136.14 = 0.0245

To get the proportion you have to divide by the least amount, in this case 0.0245

water = 0.04896 / 0.0245 = 2

calcium sulphate = 0.0245 / 0.0245 = 1

Therefore the formula is CaSO₄ · 2 H₂O, X = 2

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The question is incomplete, here is the complete question:

Sulfuric acid is essential to dozens of important industries from steel making to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds 2.0×10¹¹ kg per year.

The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 1.8 kg of solid sulfur and 10.0 atm of oxygen gas at 650°C into an evacuated 50.0 L tank. The engineer believes Kp = 0.099 for the reaction at this temperature.

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<u>Explanation:</u>

The chemical equation for the formation of sulfur dioxide gas follows:

$S(s)+O_2\rightarrow SO_2(g)$

<u>Initial:</u> 10.0

<u>At eqllm:</u> 10-x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{SO_2}}{p_{O_2}}$

We are given:

$K_p=0.099$

Putting values in above expression, we get:

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To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the sulfur dioxide gas = 0.901 atm

V = Volume of the gas = 50.0 L

T = Temperature of the gas = $650^oC=[650+273]K=923K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of sulfur dioxide gas = ?

Putting values in above equation, we get:

$0.901atm\times 50.0L=n\times 0.0821\text{ L. atm}mol^{-1}K^{-1}\times 923K\\\\n=\frac{0.901\times 50.0}{0.0821\times 923}=0.594mol$

By stoichiometry of the reaction:

1 mole of sulfur dioxide gas is produced from 1 mole of sulfur

So, 0.594 moles of sulfur dioxide gas will be produced from = $\frac{1}{1}\times 0.594=0.594mol$ of sulfur

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of sulfur = 0.594 moles

Molar mass of sulfur = 32 g/mol

Putting values in above equation, we get:

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