Question

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)

Answer 1

Answer: The mass of $AgC_{10}H_9N_4SO_2$ produced is 21.13 grams

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For silver (I) oxide:

Given mass of silver (I) oxide = 3.024 g

Molar mass of silver (I) oxide = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of silver (I) oxide}=\frac{3.024g}{102.1g/mol}=0.0296mol$

• For $C_{10}H_{10}N_4SO_2$ :

Given mass of $C_{10}H_{10}N_4SO_2$ = 50.0 g

Molar mass of $C_{10}H_{10}N_4SO_2$ = 250 g/mol

Putting values in equation 1, we get:

$\text{Moles of }C_{10}H_{10}N_4SO_2=\frac{50.0g}{250g/mol}=0.2mol$

The chemical equation for the combustion of hexane follows:

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

By Stoichiometry of the reaction:

1 mole of silver (I) oxide reacts with 2 moles of $C_{10}H_{10}N_4SO_2$

So, 0.0296 moles of silver (I) oxide will react with = $\frac{2}{1}\times 0.0296=0.0592mol$ of $C_{10}H_{10}N_4SO_2$

As, given amount of $C_{10}H_{10}N_4SO_2$ is more than the required amount. So, it is considered as an excess reagent.

Thus, silver (I) oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of silver (I) oxide produces 2 moles of $AgC_{10}H_9N_4SO_2$

So, 0.0296 moles of silver (I) oxide will produce = $\frac{2}{1}\times 0.0296=0.0592moles$ of $AgC_{10}H_9N_4SO_2$

Now, calculating the mass of $AgC_{10}H_9N_4SO_2$  from equation 1, we get:

Molar mass of $AgC_{10}H_9N_4SO_2$  = 357 g/mol

Moles of $AgC_{10}H_9N_4SO_2$  = 0.0592 moles

Putting values in equation 1, we get:

$0.0592mol=\frac{\text{Mass of }AgC_{10}H_9N_4SO_2}{357g/mol}\\\\\text{Mass of }AgC_{10}H_9N_4SO_2=(0.0592mol\times 357g/mol)=21.13g$

Hence, the mass of $AgC_{10}H_9N_4SO_2$ produced is 21.13 grams