Question 6 of 10

What volume (in liters, at 703 k and 2.04 atm) of chlorine gas is required to react with 3.39 g of p?

Natali5045456 [20]

The volume of chlorine required is 7.71 L.

The reaction between phosphorus and chlorine is:

2P + 5Cl₂→ 5PCl₅

Therefore, 2 moles of P requires 5 moles of chlorine to react with it.

Given mass of P =3.39 g

Molar mass of P=30.97 g/mol

No. of moles of P = given mass/ molar mass = 3.39 / 30.97 = 0.109 moles

2 moles of P requires 5 moles of chlorine

0.109 moles of P will require 0.109 x 5/2 = 0.2725 moles of chlorine

According to ideal gas equation

PV=nRT

2.04 x V = 0.2725 x 0.0821 x 703

V = 0.2725 x 0.0821 x 703 / 2.04

V = 7.71L

Learn more about ideal gas equation here:

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5 0

2 years ago

he chlorination of methane occurs in a number of steps that results in the formation of chloromethane and hydrogen chloride. The

frez [133]

Answer:

The total pressure in the flask is 0,619 atm.

Explanation:

For the reaction:

2CH₄(g) + 3Cl₂(g) → 2 CH₃Cl(g) + 2HCl(g) + 2Cl⁻(g)

The moles of CH₄ in 295 mL at STP are:

n = PV/RT

Where P is pressure (1 atm), V is volume (0,295L), R is gas constant (0,082atmL/molK) and T is temperature (273,15 K)

Replacing, moles of CH₄ are 0,0132 moles

In the same way, moles of chlorine are 0,0324 moles

As 3 moles of Cl₂ react with 2 moles of CH₄, for a total reaction of 0,0132 moles of CH₄ you need:

0,0132 moles CH₄ × $\frac{3 mol Cl_{2}}{2 mol CH_{4}}$ = 0,0198 moles Cl₂. That means that 0,0324-0,0198 = -0,0126 moles of Cl₂ are in excess-.

As the reaction reaches in 77%, the moles of CH₄ that don't react are:

0,0132×(100%-77%)= 3,036x10⁻³ moles of CH₄

Also, the moles of Cl₂ that don't react are:

0,0126 + 0,0198×(100%-77%)= 0,0172 moles of Cl₂

The moles produced of each compound are:

0,0132×77% moles × $\frac{2 mol CH_{3}Cl}{2 mol CH_{4}}$ = 0,0102 moles of CH₃Cl -that are the same moles of HCl and Cl⁻-

Thus, total moles in the flask are:

3,036x10⁻³ moles of CH₄ + 0,0172 moles of Cl₂ + 0,0102 moles of CH₃Cl + 0,0102 moles of HCl + 0,0102 moles of Cl⁻ = 0,0507 total moles

As the volume of the flask is 2,00L and the final temperature is 298 K. The total pressure in the flask is:

P = nRT/V

P = 0,619 atm

I hope it helps!

5 0

3 years ago

When it's freezing outside, how does this insulation prevent the pipes from bursting?

xxTIMURxx [149]

Answer:

let the cold water trip from the faucet

8 0

3 years ago

Write the equation for the formation of water gas

GREYUIT [131]

Answer:

the answer is C + H2O # CO + H2.

4 0

3 years ago

Read 2 more answers

Calculate Δ H° for the reaction C 4H 4( g) + 2H 2( g) → C 4H 8( g), using the following data: Δ H° combustion for C 4H 4( g) = –

arlik [135]

Answer: The enthalpy of the reaction is coming out to be 2231 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_{(product)}]-\sum [n\times \Delta H^o_{(reactant)}]$

For the given chemical reaction:

$C_4H_4(g)+2H_2(g)\rightarrow C_4H_8(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_{(C_4H_8(g))})]-[(1\times \Delta H^o_{(C_4H_4(g))})+(2\times \Delta H^o_{(H_2(g))})]$

We are given:

$\Delta H^o_{(C_4H_8(g))}=-2755kJ/mol\\\Delta H^o_{(H_2(g))}=-286kJ/mol\\\Delta H^o_{(C_4H_4(g))}=-2341kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-2755))]-[(1\times (-286))+(2\times (-2341))]\\\\\Delta H^o_{rxn}=2213kJ$

Hence, the enthalpy of the reaction is coming out to be 2231 kJ.

5 0

3 years ago