________ properties depend on the amount of material present.

2 H2S(g) ⇄ 2 H2(g) + S2(g) Kc = 9.3× 10-8 at 400ºC 0.47 moles of H2S are placed in a 3.0 L container and the system is allowed t

Anon25 [30]

Answer: The concentration of hydrogen gas at equilibrium is $1.648\times 10^{-3}M$

Explanation:

We are given:

Initial moles of hydrogen sulfide gas = 0.47 moles

Volume of the container = 3.0 L

The molarity of solution is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of the solution}}$

So, $\text{Initial molarity of hydrogen sulfide gas}=\frac{0.47}{3}=0.1567M$

The given chemical equation follows:

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

Initial: 0.1567

At eqllm: 0.1567-2x 2x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

We are given:

$K_c=9.3\times 10^{-8}$

Putting values in above equation, we get:

$9.3\times 10^{-8}=\frac{(2x)^2\times x}{(0.1567-2x)^2}\\\\x=8.24\times 10^{-4}$

So, equilibrium concentration of hydrogen gas = $2x=(2\times 8.24\times 10^{-4})=1.648\times 10^{-3}M$

Hence, the concentration of hydrogen gas at equilibrium is $1.648\times 10^{-3}M$

7 0

3 years ago

What is the periodic law

Schach [20]

A law stating that the elements are listed in the order of their atomic number

5 0

3 years ago

At 45 C the volume of a certain gas is 27.5 L

wariber [46]

Regretfully, I do not know what is asked of the problem because it is not stated above. However, it is easier to answer any question pertaining to the temperature, volume, and pressure of gases by assuming that they behave ideally. With this, we'll be able to use the ideal gas law which is mathematically expressed as,
PV = nRT
where P and V are pressure and volume, respectively. n is the amount of material in mol, R is gas constant, and T is temperature.

4 0

4 years ago

Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g) Which half reaction correctly describes the oxidation that is taking place?

abruzzese [7]

Zn (s) -> Zn+2 (aq) + 2e-

Zn (s) with a neutral charge is oxidized and looses two electrons in the process to form ZnCl2 (aq) where Zn has a charge of 2+.

8 0

4 years ago

Read 2 more answers

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

MAVERICK [17]

Answer: The percent yield of water is 46.9 %

Explanation:

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

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

For sulfuric acid:

Given mass of sulfuric acid = 72.6 g (Assuming)

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol$

For NaOH:

Given mass of NaOH = 77 g (Assuming)

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{77g}{40g/mol}=1.925mol$

The chemical equation for the reaction of sulfuric acid and sodium hydroxide follows:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of NaOH

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

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

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of water

0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

$1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.67g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 12.5 g (Assuming)

Theoretical yield of water = 26.67 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{12.5g}{26.67g}\times 100\\\\\% \text{yield of water}=46.9\%$

Hence, the percent yield of water is 46.9 %

3 0

3 years ago