I'll give brianliest if correct .

A 6.00 L vessel contains 20.0 g of PCl3 and 3.15 g of O2 at 15.0 ∘C. The vessel is heated to 210 ∘C, and the contents react to g

dusya [7]

<u>Answer:</u> The final pressure in the vessel will be 0.965 atm

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For phosphorus trichloride:</u>

Given mass of phosphorus trichloride = 20.0 g

Molar mass of phosphorus trichloride = 137.3 g/mol

Putting values in equation 1, we get:

$\text{Moles of phosphorus trichloride}=\frac{20.0g}{137.3g/mol}=0.146mol$

<u>For oxygen gas:</u>

Given mass of oxygen gas = 3.15 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{3.15g}{32g/mol}=0.098mol$

The chemical equation for the reaction of phosphorus trichloride and oxygen gas follows:

$2PCl_3+O_2\rightarrow 2POCl_3$

By Stoichiometry of the reaction:

2 moles of phosphorus trichloride reacts with 1 mole of oxygen gas

So, 0.146 moles of phosphorus trichloride will react with = $\frac{1}{2}\times 0.146=0.073mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, phosphorus trichloride is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of phosphorus trichloride produces 2 moles of $POCl_3$

So, 0.146 moles of phosphorus trichloride will produce = $\frac{2}{2}\times 0.146=0.146mol$ of $POCl_3$

To calculate the pressure of the vessel, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the vessel = ?

V = Volume of the vessel = 6.00 L

T = Temperature of the vessel = $210^oC=[210+273]K=483K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles = 0.146 moles

Putting values in above equation, we get:

$P\times 6.00L=0.146mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 483K\\\\P=\frac{0.146\times 0.0821\times 483}{6.00}=0.965atm$

Hence, the final pressure in the vessel will be 0.965 atm

3 0

3 years ago

Read 2 more answers

When a hydrocarbon fuel is burned, almost all of the carbon in the fuel burns completely to form co2 (carbon dioxide), which is

xxTIMURxx [149]

<h3>Answer:</h3>

1.5 × 10⁸ Kg of CO₂ / Year

<h3>Solution:</h3>

According to statement,

1 kwh generation produces = 0.73 Kg of CO₂

So,

700 kwh generation will produce = X Kg of CO₂

Solving for X,

X = (700 kwh × 0.73 Kg) ÷ 1 kwh

X = 511 Kg of CO₂ / Year

Also, as calculated,

1 household produces = 511 Kg of CO₂ / Year

Then,

300,000 households will produce = X Kg of CO₂ per Year

Solving for X,

X = (300000 households × 511 Kg) ÷ 1 household

X = 153300000 Kg of CO₂ / Year

Or,

X = 1.5 × 10⁸ Kg of CO₂ / Year

6 0

3 years ago

What is the percent composition of iron in iron III chloride?

lara31 [8.8K]

The correct answer is c

7 0

3 years ago

How do you solve for limiting reagent with a balanced chemical equation? What are the steps?

Serggg [28]

Balance first the chemical equation given if it's unbalance before moving on to first step of mathematical computation.

Download pdf

6 0

3 years ago

How many moles of air must escape from a 10-m  8.0-m  5.0-m room when the temperature is raised from 0c to 20c? assume the p

slega [8]

Data:

p = 1 atm
V = 10 m * 8 m * 5 m = 400 m^3 = 400,000 liter

To = 0 + 273.15K = 273.15K
Tf = 20 + 273.15K = 293.15K

No - Nf =?

2) Formula

pV = NRT => N = pV / (RT)

3) solution

No = pV / (RTo)

Nf = pV / (RTf)

=> No - Nf = [pv / R] [ 1 / To - 1 / Tf ]

=> No - Nf = [1atm*400,000liter / 0.0821 atm*liter/K*mol ] [ 1 / 273.15 - 1 / 293.15]

No - Nf = 1216.9 moles ≈ 1217 moles

Answer: 1217 moles

3 0

3 years ago