The reaction N2 + 3 H2 → 2 NH3 is used to produce ammonia. When 450. g of hydrogen was reacted with nitrogen, 1575 g of ammonia were produced. What is the percent yield of this …

Question

asked Dec 14, 2020 186k views

1 Answer

Derek Morrison · Answer 1 · 2020-12-21T09:05:05+0000

Answer : The percent yield of the reaction is, 61.8 %

Solution : Given,

Mass of
H_2 = 450 g

Molar mass of
H_2 = 2 g/mole

Molar mass of
NH_3 = 17 g/mole

First we have to calculate the moles of
.

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=(450g)/(2g/mole)=225moles$

Now we have to calculate the moles of

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction, we conclude that

As, 3 mole of
H_2 react to give 2 mole of
NH_3

So, 225 moles of
H_2 react to give
(225)/(3)* 2=150 moles of
NH_3

Now we have to calculate the mass of

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

$\text{ Mass of }NH_3=(150moles)* (17g/mole)=2550g$

Theoretical yield of
NH_3 = 2550 g

Experimental yield of
NH_3 = 1575 g

Now we have to calculate the percent yield of the reaction.

$\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theretical yield of }NH_3}* 100$

$\% \text{ yield of }NH_3=(1575g)/(2550g)* 100=61.8\%$

Therefore, the percent yield of the reaction is, 61.8 %