Chemistry, 04.08.2019 05:00 esme72

A2.00-l reaction vessel, initially at 298 k, contains chlorine gas at a partial pressure of 337 mmhg and fluorine gas at a partial pressure of 729 mmhg. identify the limiting reactant and determine the theoretical yield of clf3 in grams.

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Answer from: mattydoug4818

The theoretical yield of ClF_3 is 2.24 g

Explanation:

The equation given by ideal gas follows:

PV=nRT .......(1)

where, P = pressure of the gas

V = Volume of the gas

T = Temperature of the gas

R = Gas constant = $62.3637\text{ L mmHg }mol^{-1}K^{-1}$

n = number of moles of gas

For fluorine gas:

We are given:

$P=337mmHg\\V=2.00L\\T=298K\\n=?$

Putting values in equation 1, we get:

$337mmHg\times 2.00L=n\times 62.3637\text{ L mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{337\times 2.00}{62.3637\times 298}=0.0363mol$

For chlorine gas:

We are given:

$P=729mmHg\\V=2.00L\\T=298K\\n=?$

Putting values in equation 1, we get:

$729mmHg\times 2.00L=n\times 62.3637\text{ L mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{729\times 2.00}{62.3637\times 298}=0.0784mol$

The chemical equation for the reaction of chlorine gas and fluorine gas follows:

$Cl_2+3F_2\rightarrow 2ClF_3$

By Stoichiometry of the reaction:

3 moles of fluorine gas reacts with 1 mole of chlorine gas.

So, 0.0363 moles of fluorine gas will react with = $\frac{1}{3}\times 0.0363=0.0121mol$ of chlorine gas

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

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

By Stoichiometry of the reaction:

3 moles of fluorine gas produces 2 moles of ClF_3

So, 0.0363 moles of fluorine gas will produce = $\frac{2}{3}\times 0.0363=0.0242mol$ of ClF_3

To calculate the theoretical yield of ClF_3

we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of ClF_3 = 92.45 g/mol

Moles of ClF_3 = 0.0242 moles

Putting values in above equation, we get:

$0.0242mol=\frac{\text{Mass of }ClF_3}{92.45g/mol}\\\\\text{Mass of }ClF_3=(0.0242mol\times 92.45g/mol)=2.24g$

Hence, the theoretical yield of ClF_3 is 2.24 g

Answer from: Quest

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

Answer from: Quest

They can measure the amount of energy gained or lost

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A2.00-l reaction vessel, initially at 298 k, contains chlorine gas at a partial pressure of 337 mmhg...

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