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3 years ago

HELPP nowww plssss!!!!

Chemistry

1 answer:

yKpoI14uk [10]3 years ago

4 0

The answer is #2: gas

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Part 1 : Balanced reaction, $3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Part 2 : The theoretical yield of $NH_3$ gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of $N_2$ = 475 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $NH_3$ = 17 g/mole

Experimental yield of $NH_3$ = 397 g

Answer for Part (1) :

The balanced chemical reaction is,

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Answer for Part (2) :

First we have to calculate the moles of $N_2$ .

$\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles$

From the given reaction, we conclude that

1 moles of $N_2$ gas react to give 2 moles of $NH_3$ gas

16.96 moles of $N_2$ gas react to give $\frac{2}{1}\times 16.96=33.92$ moles of $NH_3$ gas

Now we have to calculate the mass of $NH_3$ gas.

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g$

Therefore, the theoretical yield of $NH_3$ gas = 440.96 g

Answer for Part (3) :

Formula used for percent yield :

$\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100$

$\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%$

Therefore, the % yield of ammonia is 90.03 %

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