If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced as well as the total number of moles of gas present at the conclusion of the reaction.

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

P₄: 1 mole
H₂: 6 moles
PH₃:4 moles

Being the molar mass of the compounds:

P₄: 124 g/mole
H₂: 2 g/mole
PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

P₄: 1 mole* 124 g/mole= 124 g
H₂: 6 mole* 2 g/mole= 12 g
PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

P= 2 atm
V= 10 L
n= ?
R= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]
T= 298 K

Replacing:

2 atm*10 L= n*0.082 [tex]\frac{atm*L}{mol*K}[/tex] *298 K

and solving you get:

[tex]n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }[/tex]

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

[tex]mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}[/tex]

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

[tex]moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3} }{124grams of P_{4}}[/tex]

moles of PH₃=0.2742

If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

[tex]moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2} }{124grams of P_{4}}[/tex]

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Oeerivonaid Oeerivonaid · Answer 2

The ideal gas equation gives the approximate amount of PH₃ produced

and the gas at the conclusion as 0.28 and 0.69 moles respectively.

Response:

First part:

The PH₃ gas produced is approximately 0.28 moles

Second part:

The number of moles of gas present at the conclusion of the reaction is approximately 0.69 moles

What is the ideal gas equation, and how does it give the number of moles of gas present?

Given:

Mass of phosphorus = 8.50 g

Pressure of hydrogen gas, P = 2.00 atm

Volume of hydrogen gas, V = 10.0–L

Temperature of the gas, T = 298 K

First part:

From the ideal gas equation, P·V = n·R·T, we have;

[tex]Number \ of \ moles, \ n = \mathbf{\dfrac{P\cdot V}{R \cdot T}}[/tex]

Where for the hydrogen gas, H₂, we have;

P = 2.00 atm

V = 10.0 L

T = 298 K

The universal gas constant, R = 0.08205 L·atm·mol⁻¹·K⁻¹

Which gives;

[tex]n = \mathbf{ \dfrac{2.00 \times 10.0}{0.08205 \times 298}} \approx 0.82[/tex]

The number of moles of H₂ in the reaction, n ≈ 0.82 moles

The equation of the reaction is; P₄(s) + 6H₂(g) [tex]\longrightarrow[/tex] 4PH₃(g)

Therefore;

1 mole of P₄(s) and 6 moles of H₂(g) produces 4 moles of PH₃(g)

Molar mass of P₄ ≈ 123.9 g

Which gives;

Number of moles in 8.50 g of phosphorus = [tex]\frac{8.5}{123.9}[/tex] ≈ 0.069

0.069 moles of P₄ will react with 0.069 × 6 moles ≈ 0.41 moles of H₂ to produce 0.069 × 4 moles ≈ 0.28 moles of PH₃

Number of moles of PH₃ gas produced is approximately 0.28 moles

Second part:

The total number of moles of gas present at the conclusion of the reaction is therefore;

∑Gas ≈ 0.28 moles + (0.82 - 0.41) moles ≈ 0.69 moles

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