Q21 P 1.475-g sample containing NH4Cl... [FREE SOLUTION]

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Chapter 27: Q21 P (page 768)

1.475-g sample containing ${NH}_{4} Cl (FM 53.491), K_{2} {CO}_{3}$ (FM 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0-mL aliquot was acidified and treated with excess sodium tetraphenylborate, ${Na}^{+} B (C_{6} H_{5})_{4}^{-}$ , to precipitate $K^{+}$ and
${NH}_{4}^{+}$ ions completely:
$(C_{6} H_{5})_{4} B^{-} + K^{+} 鈫� (C_{6} H_{5})_{4} B K (s) F M 358.33 (C_{6} H_{5})_{4} B^{-} + N H_{4}^{+} 鈫� (C_{6} H_{5})_{4} B N H_{4} (s) F M 337.27$
The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original solution was made alkaline and heated to drive off all the ${NH}_{3} .$
$N H_{4}^{+} + O H^{-} 鈫� N H_{3} (g) + H_{2} O$
It was then acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. Find the weight percent of ${NH}_{4} Cl$ and $K_{2} {CO}_{3}$ in the original solid.

Short Answer

Expert verified

The weight percentage of $m ({NH}_{4} Cl)$ is 14.6%

The weight percentage of $m (N_{2} {CO}_{3})$ is 14.5%

Step by step solution

Calculate x and y:

Let consider,

X as $m ({NH}_{4} Cl)$ and y as $m (K_{2} C O_{3})$

25 ml of the sample gave 0.617 g of precipitate which contained both of the products

$\frac{1}{4} (\frac{x}{M (N H_{4} C l)}) 脳 M ((C_{6} H_{5})_{4} B N H_{4}) + (\frac{(2 y)}{M (K_{2} C O_{3})} 脳 M ((C_{6} H_{5})_{4} B K)) = 0.617 g \frac{1}{4} (\frac{x}{53.492} 脳 337.27 + \frac{2 y}{138.21} 脳 358.33) = 0.617 g$

Calculate for y we get,

$\frac{1}{2} (\frac{2 y}{M (K_{2} C O_{3})} 脳 M ((C_{6} H_{5})_{4} B K))) = 0.554 g \frac{1}{2} (\frac{2 y}{138.21} 脳 358.33) = 0.554 g y = 0.2137 g$

Calculate for x we get,

$\frac{1}{4} (\frac{x}{53.492} 脳 337.27 + \frac{2 脳 0.2137 g}{138.21} 脳 358.33) = 0.617 g x = 0.2157 g$

Calculate the weight percentages:

$w t % (x) = \frac{x}{m (s a m p l e)} w t % (x) = \frac{0.2157 g}{1.475 g} w t % (x) = 14.6 %$

The weight percentage of $m ({NH}_{4} Cl)$ is 14.6%

role="math" localid="1663664775980" $w t % (y) = \frac{y}{m (s a m p l e)} w t % (y) = \frac{0.2137 g}{1.475 g} w t % (y) = 14.5 %$

The weight percentage of $m (K_{2} C O_{3})$ is 14.5%

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Short Answer

Step by step solution

Calculate x and y:

Calculate the weight percentages:

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