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3 years ago

Calculate percent composition of a compound that forms when 222.6g n combines 77.4g o

1 answer:

6 0

It must be noted that our compound contains only two components, N and O. From the given masses, the total mass of the compound should be 300 g.
Then, we divide the masses of each component by the total mass and multiply the quotients by 100% to get the % compositions
N = (222.6 g / 300 g) x 100% = 74.2%
O = (77.4 g / 300 g) x 100% = 25.8%

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Which element has four times as many protons in its nucleus than are found in Neon?

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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)

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Answer: 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.

Explanation:

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:

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

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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.

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