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kari74 [83]
3 years ago
5

If the reaction below takes place at STP how many L of ammonia (NH3) will be formed if one starts with 3.45 moles of hydrogen an

d an excess of nitrogen? Be sure to balance the equation before you begin and round to the proper accuracy. Please leave any units off.
Chemistry
1 answer:
lesya692 [45]3 years ago
6 0

Answer:

51.5 L of ammonia

Explanation:

Ammonia is produced according to this equation:

N₂(g) + 3H₂(g) → 2NH₃ (g)

Let's propose this rule of three:

3 moles of hydrogen can produce 2 moles of ammonia

Then 3.45 moles of hydrogen, will produce (3.45 . 2) / 3 = 2.3 moles of NH₃

We apply the Ideal Gases Law at STP to determine the volume

STP = 1 atm and 273K

1 atm . V = 2.3 mol . 0.082 . 273K

V = (2.3 mol . 0.082 . 273K) / 1 atm = 51.5 L

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The colligative molality of an unknown aqueous solution is 1.56 m.
yawa3891 [41]

Answer:

Vapor pressure of solution = 17.02 Torr

T° of boiling point for the solution is 100.79°C

T° of freezing point for the solution is -2.9°C

Explanation:

Let's state the colligative properties with their formulas

- <u>Vapor pressure lowering</u>

ΔP = P° . Xm . i

- <u>Boiling point elevation</u>

ΔT = Kb . m . i

-<u> Freezing point depressión</u>

ΔT = Kf . m . i

ΔP = Vapor pressure pure solvent (P°) - Vapor pressure solution

ΔT = T° boling solution - T° boiling pure solvent

ΔT = T° freezing pure solvent - T° freezing solution

i represents the Van't Hoff factor (ions dissolved in the solution). If we assume that the solute is non-volatile and the solution is ideal i = 1

Kf and Kb are cryoscopic and ebulloscopic constant, they are  specific to each solvent.

Vapor pressure works with mole fraction (Xm) and the only data we have is molality, so we consider 1.56 moles of solute and 1000 g of solvent mass.

Moles of solvent → solvent mass / molar mass of solvent

Moles of solvent → 1000 g / 18 g/mol = 55.5 moles

Mole fraction is moles of solute / Total moles (mol st + mol sv)

Mole fraction: 1.56 / (1.56 + 55.5) = 0.027

- Vapor pressure lowering

ΔP = P° . Xm . i

17.5 Torr - Vapor pressure of solution = 17.5 Torr . 0.027 . 1

Vapor pressure of solution = - (17.5 Torr . 0.027 . 1 - 17.5 Torr)

Vapor pressure of solution = 17.02 Torr

- Boiling point elevation

ΔT = Kb . m . i

T° boiling solution - 100° = 0.512 °C/ m . 1.56 m . 1

T°boiling solution = 0.512 °C/ m . 1.56 m . 1 + 100°C

T°boiling solution = 100.79°C

- Freezing point depression

ΔT = Kf . m . i

0°C - T° freezing solution = 1.86 °C/m . 1.56 m . 1

T° freezing solution = - (1.86 °C/m . 1.56 m)

T° freezing solution = -2.9°C

3 0
3 years ago
N2 + 3H2, -&gt; 2NH3<br> If I have 50.6 L of N2 and excess H2, how many liters of NH3 can I produce?
Anastaziya [24]

Answer:

C. 101.2 L

Explanation:

N2 + H2= NH3

Balancing it,

N2 + 3 H2 = 2.NH3

(1 mol) (3 mol) (2 mol)

which means

1 molecule of nitrogen reacts with 3 molecule of hydrogen to form ammonia.

Likewise,

50.6 l of nitrogen reacts with 50.6 × 3= 151.8 l of hydrogrn to form 50.6 × 2= 101.2 l of ammonia.

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New pressure would be 6.36 kPa
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Answer:

a) 30.97

b) 3s² 3p²

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