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Hoochie [10]
4 years ago
8

How to writhe the balanced chemical equation

Chemistry
1 answer:
mariarad [96]4 years ago
3 0
Get an equal amount of each element by adding coefficients to the different compounds.
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In a laboratory setting, chemists strive to be
Kitty [74]
Accurate and precise
3 0
4 years ago
Why is it easier to float in the sea than freshwater?
natali 33 [55]
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5 0
4 years ago
The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: CO(g) + H2O(g)
taurus [48]

<u>Answer:</u> The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

<u>Explanation:</u>

For the given chemical reaction:

CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

The expression of K_c for above reaction follows:

K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}         ........(1)

We are given:

[CO]_{eq}=[H_2O]_{eq}=[H_2]_{eq}=0.10M

[CO_2]_{eq}=0.40M

Putting values in above equation, we get:

K_c=\frac{0.40\times 0.10}{010\times 0.10}\\\\K_c=4

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of hydrogen gas = 0.30 mol

Volume of solution = 2.0 L

Putting values in above equation, we get:

\text{Molarity of }H_2=\frac{0.30mol}{2L}=0.15M

When hydrogen gas is added, the concentration of product gets increased. But, by Le-Chatelier's principle, the equilibrium will shift in the direction where concentration of product must decrease, which is in the backward direction.

Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M

Now, the equilibrium is shifting to the reactant side. The equation follows:

                      CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial:              0.1      0.1                 0.4       0.1

At eqllm:       0.1+x   0.1+x           0.4-x      0.25-x

Putting values in expression 1, we get:

4=\frac{(0.25-x)(0.4-x)}{(0.1+x)(0.1+x)}\\\\3x^2+1.45x-0.06=0\\\\x=0.038,-0.522

Neglecting the negative value of 'x'

Calculating the concentrations of the species:

Concentration of carbon dioxide = (0.4 - x) = (0.4 - 0.038) = 0.362 M

Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M

Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Hence, the concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

8 0
4 years ago
For the following reaction, 35.4 grams of zinc oxide are allowed to react with 6.96 grams of water . zinc oxide(s) + water(l) --
IRISSAK [1]

Answer:

m_{Zn(OH)_2}=38.4g

Explanation:

Hello!

In this case, for the undergoing chemical reaction:

ZnO(s)+H_2O(l)\rightarrow Zn(OH)_2

We evaluate the yielded moles of zinc hydroxide by each reactant as shown below:

n_{Zn(OH)_2}^{by ZnO}=35.4gZnO*\frac{1molZnO}{81.38gZnO}*\frac{1molZn(OH)_2}{1molZnO}  =0.435molZn(OH)_2\\\\n_{Zn(OH)_2}^{by H_2O}=6.96gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molZn(OH)_2}{1molH_2O}  =0.386molZn(OH)_2

In such a way, since the water yields a smaller amount of zinc hydroxide we conclude it is the limiting reactant so the maximum mass is computed below:

m_{Zn(OH)_2}=0.386molZn(OH)_2*\frac{99.424 gZn(OH)_2}{1molZn(OH)_2} \\\\m_{Zn(OH)_2}=38.4g

Because the water limits the yielded amount of zinc hydroxide.

Best regards!

5 0
3 years ago
Calculate the mass of 29.6 L of fluorine gas at STP.
Lunna [17]

Answer:

Mass = 25.08 g

Explanation:

Given data:

Volume of fluorine = 29.6 L

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Mass of fluorine = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

n = PV/RT

n = 1 atm × 29.6 L / 0.0821 atm.L/ mol.K × 273.15 K

n = 1.32 mol

Mass of fluorine:

Mass = number of moles × molar mass

Mass = 1.32 mol ×19 g/mol

Mass = 25.08 g

7 0
3 years ago
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