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3 years ago

What is the density of ammonia (NH3) at 293 K and 0.913 atm?

Chemistry

1 answer:

Anastaziya [24]3 years ago

7 0

Answer: The density of Ammonia is 0.648 g/l

Explanation:

Density = Mass/ Volume

Mass of one mole of Ammonia (NH3) = 17.031g

Volume =?

Using the ideal gas law we can determine the volume.

PV = nRT

P = 0.913 atm, V= ?, n = 1, R = 0.08206 L.atm/K, and T= 293K

Make V the subject of the formular, we then have;

V= nRT/ P = 1 mol x 0.08206 L.atm/ K.mol x 293 / 0.913 atm

V = 24.04358/ 0.913 = 26.3L

Having gotten the value of Volume in this question, we then go back to solve for density.

Density = Mass/ Volume

17.031g/ 26.3L = 0.64756 ≈ 0.648 g/l

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Answer:

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It is given that,

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How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

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4 years ago