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3 years ago

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What mass of silver oxide, Ag2O is required to produce 25.0 g of silver sulfadiazine, AgC10H9N4SO2, from the reaction of silver

oxide and sulfadiazine?
2C10H10N4SO2 + Ag2O ? 2AgC10H9N4SO2 + H2O

1 answer:

Wittaler [7]3 years ago

4 0

Answer:

8.1107 g

Explanation:

The given reaction:

$2C_{10}H_{10}N_4SO_2+Ag_2O\rightarrow 2AgC_{10}H_9N_4SO_2+H_2O$

Given that:

Mass of silver sulfadiazine = 25.0 g

Molar mass of silver sulfadiazine = 357.14 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{25.0\ g}{357.14\ g/mol}$

$Moles= 0.07\ mol$

From the reaction,

2 moles of silver sulfadiazine are formed from 1 mole of silver oxide

So,

1 mole of silver sulfadiazine are formed from 1/2 mole of silver oxide

0.07 mole of silver sulfadiazine are formed from 1/2*0.07 mole of silver oxide

Moles of silver oxide = 0.035 moles

Molar mass of silver oxide = 231.735 g/mol

Mass = Moles * Molar mass = 0.035 moles * 231.735 g/mol = 8.1107 g

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Approximately $81.84\%$ .

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Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

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$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

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