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Anna007 [38]
4 years ago
7

The reaction for the decomposition of ammonia (NH3) can be written as shown. If a student starts with 21.7 g of ammonia, how man

y grams of hydrogen gas (H2) will be produced by the reaction?
Chemistry
2 answers:
Eddi Din [679]4 years ago
3 0
3.88 grams will be released I believe
Allisa [31]4 years ago
3 0

Answer:- 3.83 g of hydrogen gas are formed.

Solution:- It is a stoichiometry problem and could easily be solved using dimensional analysis. The balanced equation for the decomposition of ammonia gas to give nitrogen and hydrogen gases is:

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

From balanced equation, there is 2:3 mol ratio between ammonia and hydrogen. We start with given grams of ammonia and convert them to moles on dividing the grams by molar mass.

In next step the moles of ammonia are multiplied by mol ratio to get the moles of hydrogen which are finally multiplied by it's molar mass to get the grams of hydrogen gas formed.

The set is shown below:

21.7gNH_3(\frac{1mol NH_3}{17g NH_3})(\frac{3mol H_2}{2mol NH_3})(\frac{2g H_2}{1mol H_2})

= 3.83g H_2

So, from the calculations 3.83 g of hydrogen gas are formed by the decomposition of 21.7 g of ammonia gas.

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The following question appears on a quiz: ""You fill a tank with gas at 60°C to 100 kPa and seal it. You decrease the temperatu
dusya [7]

Answer: The final pressure will decrease ad the value is 85 kPa

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=100kPa\\T_1=60^0C=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K

Putting values in above equation, we get:

\frac{100kPa}{333K}=\frac{P_2}{283K}\\\\P_2=85kPa

Hence, the final pressure will decrease ad the value is 85 kPa

8 0
3 years ago
When CO2(g) is put in a sealed container at 730 K and a pressure of 10.0 atm and is heated to 1420 K , the pressure rises to 24.
d1i1m1o1n [39]

Answer:

48%

Explanation:

Based on Gay-Lussac's law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:

P1T2 = P2T1

<em>Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:</em>

P1 = 10.0atm

T2 = 1420K

P2 = ?

T1 = 730K

P2 = 10.0atm*1420K / 730K

P2 = 19.45 atm

The CO2 reacts as follows:

2CO2 → 2CO+ O2

Where 2 moles of gas react producing 3 moles of gas

Assuming the 100% of CO2 react, the pressure will be:

19.45atm * (3mol / 2mol) = 29.175atm

As the pressure rises just to 24.1atm the moles that react are:

24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present

The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:

0.48mol / 1mol * 100 = 48%

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