Question

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{g}}$of role="math" localid="1648634968074" ${\mathbf{\text{Ag}}}_{\mathbf{2}}\mathbf{\text{O}}$is reacted with$\mathbf{\text{50.0g}}$ of ${\mathit{C}}_{\mathbf{10}}{\mathit{H}}_{\mathbf{10}}{\mathit{N}}_{\mathbf{4}}\mathit{S}{\mathit{O}}_{\mathbf{2}}$, what mass of silver sulfadiazine $\left(Ag{C}_{10}{H}_9{N}_{4}S{O}_2\right)$can be produced, assuming100% yield?

${\mathbf{\text{Ag}}}_{\mathbf{2}}\mathit{O}\left(s\right)\mathbf{+}\mathbf{2}{\mathit{C}}_{\mathbf{10}}{\mathit{H}}_{\mathbf{9}}{\mathit{N}}_{\mathbf{4}}\mathit{S}{\mathit{O}}_{\mathbf{2}}\left(s\right)\mathbf{\to }\mathbf{2}\mathit{A}\mathit{g}{\mathit{C}}_{\mathbf{10}}{\mathit{H}}_{\mathbf{9}}{\mathit{N}}_{\mathbf{4}}\mathit{S}{\mathit{O}}_{\mathbf{2}}\left(s\right)\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(l\right)$

Short answer

The mass of silver sulfadiazine$\left(Ag{C}_{10}{H}_9{N}_{4}S{O}_2\right)$that can be produced is about$\mathbf{71}\mathbf{.}\mathbf{44}\mathbf{\text{g}}$ .

Step-by-step solution

Definition

Themass of any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determination of moles of  C10H10N4SO2

From the given,

$25.0\text{g}$of${\text{Ag}}_2\text{O}$ reacts with $\text{50.0g}$of ${\mathrm{C}}_{10}{\mathrm{H}}_{10}{\mathrm{N}}_4{\mathrm{SO}}_2$

${\mathbf{\text{MolesofAg}}}_{\mathbf{2}}\mathit{O}\mathbf{=}\frac{\mathbf{\text{Mass}}}{\mathbf{\text{Molarmass}}}$

Molar mass of ${\text{Ag}}_2\text{O=}\left(2\left(107.9\right)+16.00\right)\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

=$231.8\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

$=\frac{25.0\text{g}}{23.8{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}$

$=0.108\text{moles}$

Thus the molar mass$=250.29\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

Moles of ${\mathrm{C}}_{10}{\mathrm{H}}_{10}{\mathrm{N}}_4{\mathrm{SO}}_2$$=\frac{50.0g}{\text{250.29}{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}$

Chemical reaction

The given equation:

${\mathrm{Ag}}_2\mathrm{O}\left(\mathrm{s}\right)+2{\mathrm{C}}_{10}{\mathrm{H}}_9{\mathrm{N}}_4{\mathrm{SO}}_2\left(\mathrm{s}\right)\to 2{\mathrm{AgC}}_{10}{\mathrm{H}}_9{\mathrm{N}}_4{\mathrm{SO}}_2\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

We know that 1 mole of ${\text{Ag}}_2\text{O}$reacts with 2 moles of ${\mathrm{C}}_{10}{\mathrm{H}}_{10}{\mathrm{N}}_4{\mathrm{SO}}_2$to produce 2silvermoles of sulfadiazine.

Later,$0.108\text{moles}$ of ${\text{Ag}}_2\text{O}$reacts to produce$0\text{.206moles}$ .So, ${\mathrm{C}}_{10}{\mathrm{H}}_{10}{\mathrm{N}}_4{\mathrm{SO}}_2$is the limiting reactant.

Determination of mass of Silver sulfadiazine

Here 1 mole of ${\mathrm{C}}_{10}{\mathrm{H}}_{10}{\mathrm{N}}_4{\mathrm{SO}}_2$to produce 2 moles of silver sulfadiazine.

${\mathrm{C}}_{10}{\mathrm{H}}_{10}{\mathrm{N}}_4{\mathrm{SO}}_2$Forms$0.2$moles ${\mathrm{C}}_{10}{\mathrm{H}}_{10}{\mathrm{N}}_4{\mathrm{SO}}_2$

$=0.2\text{molesof}\left({\mathrm{AgC}}_{10}{\mathrm{H}}_9{\mathrm{N}}_4{\mathrm{SO}}_2\right)$

=$357.182\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Thus the mass of ${\text{AgC}}_{10}{\text{H}}_9{N}_4{\text{SO}}_2=\left(\text{moles}\right)\times \left(\text{molarmass}\right)$

role="math" localid="1648632138310" $\text{=0.2×moles×357.82}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$$=71.44g$

Therefore, the mass of ${\mathbf{\text{AgC}}}_{\mathbf{10}}{\mathbf{\text{H}}}_{\mathbf{9}}{\mathbf{\text{N}}}_{\mathbf{4}}{\mathbf{\text{SO}}}_{\mathbf{2}}$is$\mathbf{\text{71.44g}}$ .