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Archy [21]
2 years ago
7

When 28 g of nitrogen and 6 g of hydrogen react, 34 g of ammonia are produced. if 100 g of nitrogen react with 6 g of hydrogen,

how much ammonia will be produced?
Chemistry
2 answers:
nignag [31]2 years ago
8 0

Answer : The mass of NH_3 produced will be, 34 grams.

Explanation : Given,

Mass of N_2 = 100 g

Mass of H_2 = 6 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{100g}{28g/mole}=3.57moles

\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{6g}{2g/mole}=3moles

Now we have to calculate the limiting and excess reagent.

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 with 1 mole of N_2

So, the given 3 moles of N_2 react with 1 moles of H_2

From this we conclude that, N_2 is an excess reagent because the given moles are greater than the required moles and H_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3.

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

So, 3 moles of H_2 react to give \frac{3}{3}\times 2=2 moles of NH_3

Now we have to calculate the mass of NH_3.

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

\text{Mass of }NH_3=(2mole)\times (17g/mole)=34g

Therefore, the mass of NH_3 produced will be, 34 grams.

vampirchik [111]2 years ago
4 0
N2 + 3H2 --> 2NH3
When 100g of N2 , no of moles of N2= 100/(28)=3.57 mol
no. of moles of h2 = 6/(2)=3mol
Therefore h2 is limiting reagent.
no. of moles of ammonia= 3/3*2=2moles
mass of ammonia produced= 2 mol * (14+3)= 34g
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0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2  \ + \ bH_2O \ + \ cN_2

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(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

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(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

x = \frac{2a+b}{2} \\ \\ x =  \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95

Equating the coefficient of Nitrogen

c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612

Therefore, The overall  balanced combustion  reaction can now be written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

Now;  To determine the stoichiometric F/A and A/F ratios; we have:

(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\  (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\  (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562

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Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

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(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg

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(LHV)_f = (HHV)_f - \delta H_w \\ \\  (LHV)_f = (HHV)_f  - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f   = 49.9876 \ MJ/kg -  [\frac{3.8*18}{44.2}*2.258 \ MJ/kg]  \\ \\ (LHV)_f = 46.4933 \ M/kg

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