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Mice21 [21]
3 years ago
14

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly

into the wound. If 25.0 g of Ag2O is reacted with 50.0 g of C10H10N4SO2, what mass of silver sulfadiazine (AgC10H9N4SO2) can be produced, assuming 100% yield?Ag2O(s)+2C10H10N4SO2(s)⟶2AgC10H9N4SO2(s)+H2O(l)
Chemistry
1 answer:
trapecia [35]3 years ago
8 0

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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When 91.5 g of isopropyl alcohol which has an empirical formula of C3H8O is burned in excess oxygen gas, how many grams of H2O a
Kobotan [32]

Answer:

109.7178g of H2O

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

2C3H8O + 9O2 —> 6CO2 + 8H2O

Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:

Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.

Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g

Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol

Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g

From the equation,

120.19184g of C3H8O produced 144.12224g of H20.

Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O

7 0
3 years ago
3. A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of
erik [133]

Answer:

Ba\ percentage\ in\ Mass=4.8\%

Explanation:

From the question we are told that:

Mass of mixture m=3.455g

Mass of Barium m_b=0.2815g

Equation of Reaction is given as

Ba2+ + H2SO4 => BaSO4 + 2 H+

Generally the equation for Moles of Barium  is mathematically given by

Since

 Moles of Ba^{2+} = Moles of BaSO_4

Therefore

 Moles of Ba^{2+}  = \frac{mass}{molar mass of BaSO4}  

 Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol

Generally the equation for Mass of Barium  is mathematically given by

 Mass\ of\ Ba^{2+} = Moles * Molar mass of Ba^{2+}  

 Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g

Therefore

 Ba\ percentage\ in\ Mass = mass of Ba^{2+}/mass of sample * 100%    

 Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%

 Ba\ percentage\ in\ Mass=4.8\%

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Answer: Pb(NO3)2

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