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grin007 [14]
3 years ago
12

The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen

t of Tl2SO4 in the sample if 0.1824 g of TlI was recovered.
Chemistry
1 answer:
tino4ka555 [31]3 years ago
6 0

Answer: The mass percentage of Tl_2SO_4 is 5.86%

Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

Using the molar mass of Tl_2SO_4 we have:

mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g

Finally, we can use the mass percentage formula:

mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%

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1.5 M.

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Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly
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Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of Ag_2O = 25.0 g

Mass of C_{10}H_{10}N_4SO_2 = 50.0 g

Molar mass of Ag_2O = 231.7 g/mole

Molar mass of C_{10}H_{10}N_4SO_2 = 250.3 g/mole

Molar mass of AgC_{10}H_{9}N_4SO_2 = 357.1 g/mole

First we have to calculate the moles of Ag_2O and C_{10}H_{10}N_4SO_2.

\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles

\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)

From the balanced reaction we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 1 mole of Ag_2O

So, 0.1998 moles of C_{10}H_{10}N_4SO_2 react with \frac{0.1998}{2}=0.0999 moles of Ag_2O

From this we conclude that, Ag_2O is an excess reagent because the given moles are greater than the required moles and C_{10}H_{10}N_4SO_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgC_{10}H_9N_4SO_2

From the reaction, we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 2 mole of AgC_{10}H_9N_4SO_2

So, 0.1998 mole of C_{10}H_{10}N_4SO_2 react with 0.1998 mole of AgC_{10}H_9N_4SO_2

Now we have to calculate the mass of AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

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