Calculate number of particles in 46 g of Na atoms

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Answer 1

Answer: the number of particles in 46g of Sodium (Na) atoms are 2.00088735004 mol

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A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L flask and is found to contain 8.62 g of CO, 2.60 g of H2, 43.0 g of CH4, and 48.4 g of H2O.

Answers

The given question is incomplete. But the complete question is this:

A mixture of $CH_(4)$ and $H_2O$ is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L flask and is found to contain 8.62 g of CO, 2.60 g of $H_2$ , 43.0 g of $CH_(4)$ , and 48.4 g of $H_(2)O$ . Assuming that equilibrium has been reached, calculate $K_(p)$ for the reaction.

Explanation:

As the given reaction is as follows.

$CH_4 + H_2O \rightarrow CO + 3H_2$

And, we know that

No. of moles = $\frac{mass}{\text{molar mass}}$

Therefore, calculate the moles as follows.

Moles of $CH_4$ = $(43)/(16.04)$

= 2.6808 mol

Moles of $H_2O$ = $(48.4)/(18.01528)$

= 2.6866 mol

Moles of CO = $(8.62)/(28.01)$

= 0.307747 mol

Moles of $H_(2)$ = $(2.6)/(2.01588)$

= 1.2897 mol

As, we know that

Concentration = $(moles)/(volume (L))$

Given volume = 5 L

Hence, calculate the concentration of given species as follows.

Conc. of $CH_4 = (2.6875)/(5)$

= 0.5361

Conc. of $H_2O = (2.6889)/(5)$

= 0.5373

Conc. of CO = $(0.307747)/(5)$

= 0.06155

and, Conc. of $H_2 = (1.2897)/(5)$

= 0.2579

Now, expression for equilibrium constant for the given reaction is as follows.

$K_(c) = ([CO][H_2]^(3))/([CH_4][H_2O])$

Now, putting the given values into the above formula as follows.

$K_(c) = ([0.06155][0.2579]^(3))/([0.5361][0.5373])$

$K_(c) = 3.665 * 10^(-3)$

Also, we know that

$K_p = K_c * (RT)^dn$

Consider the equation

$CH_4(g) + H2O(g) \rightarrow CO(g) + 3H_2(g)$

Calculate change in moles of gas as follows.

change in gas moles (dn) = 1 + 3 - 1 - 1

dn = 2

As, $K_p = K_c * (RT)^2$

It is given that,

T = 1000 K, R = 0.0821

So,

$K_p = 3.665 * 10^(-3) * (0.0821 * 1000)^(2)$

$K_p$ = 24.70

Thus, we can conclude that value of $K_(p)$ for the reaction is 24.70.

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