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german
3 years ago
13

The following balanced equation shows the decomposition of ammonia (NH3) into nitrogen (N2) and hydrogen (H2). 2NH3 → N2 + 3H2 A

quantity of NH3 decomposes to produce 0.0351 mol N2. How many moles of H2 are produced? 0.0351 mol H2 0.0117 mol H2 0.105 mol H2
Chemistry
2 answers:
IRISSAK [1]3 years ago
8 0

0.105 mol (OPTION 3).

Triss [41]3 years ago
6 0
The decomposition of ammonia is characterized by the following decomposition equation:
                                  2NH₃<span>   →   N</span>₂  <span> +   3H</span>₂   

The mole ratio of N₂  :  H₂  is  1  :  3

    If the number of moles of N₂  =  0.0351 mol
    Then the number of moles of H₂  =  0.0351 mol  × 3
                                                         = 0.1053 mol

The number of moles of hydrogen gas produced when 0.0351 mol of Nitrogen gas is produced after the decomposition of Ammonia is  0.105 mol (OPTION 3).

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In standardizing a naoh solution a student found that 25.55cm of base neutralize exactly 21.35cm of 0.12M HCl find the molarity
nalin [4]

Answer:

O.1M

Explanation:

First let's generate a balanced equation for the reaction

NaOH + HCl —>NaCl + H2O

From the equation,

The ratio of the acid to base is 1:1.

From the question, we obtained the following:

Ma = Molarity of acid = 0.12M

Va = volume of acid = 21.35cm3

Vb = volume of base = 25.55cm3

Mb = Molarity of base =?

We obtained nA(mole of acid) and nB(mole of base) to be 1

The molarity of the base can be calculated for using:

MaVa/ MbVb = nA / nB

0.12x21.35 / Mb x 25.55 = 1

Cross multiply to express in linear form

Mb x 25.55 = 0.12x21.35

Divide both side by 25.55

Mb = (0.12x21.35) / 25.55

Mb = 0.1M

The molarity of the base is 0.1M

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A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

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