What element is used in bright flashing advertising signs

Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe

ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of $CH_4$ will be,

$2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)$ $\Delta H_1=29.7kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$ $\Delta H_2=-41kJ$

(3) $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ $\Delta H_3=-206kJ$

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

$\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)$

$\Delta H_{rxn}=-187.6kJ/mol$

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

$\Delta H=\frac{q}{n}$

or,

$q=\Delta H\times n$

where,

$\Delta H$ = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = $\frac{1.00\times 1000g}{12.00g/mol}=83.33mol$

Now put all the given values in the above formula, we get:

$q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ$

Thus, the total heat will be, -2251 kJ

4 0

2 years ago

If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield

Zigmanuir [339]

<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

$ActualYield=37.6g\\TheoreticalYield=112.8g$

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

$PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)$

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0

2 years ago

Read 2 more answers

1. What is the oxidation number for the silver ion in tarnish?

spayn [35]

Tarnish is Ag2S-silver sulfide and the oxidation state of silver is +1

7 0

1 year ago

True or false, The atomic number of an element is a whole number that decreases as you read across each row of

torisob [31]

Answer:

I believe it's false because the atomic number is the number of protons in the nucleus of an atom.

5 0

1 year ago

In each row, check the box under the compound that can reasonably be expected to be more acidic in aqueous solution, e.g. have t

vredina [299]

Answer:

$HCH_{3}SO_{2}$

$H_{3}PO_{3}$

$HClO_{2}$

Explanation:

Every acid (HA) tends to disolve into proton ( $H^{+}$ ) and anion ( $A^{-}$ ) in aqueous solution. Acid strength can be determined by measuring this tendency to separate into proton an anion. Strength of an acid can be quantified by its acid dissociation value - Ka. A strong acid will have a tendency to easily release proton and will have larger Ka value and smaller logarithmic value (pKa = - logKa) similar to calculating pH of the solution. So the easiest way to resolve this issue is by looking for Ka or pKa value of the acid (This table may be useful in more complex tasks and is attached below). However, stronger acid can be determined elsehow.

a) Carbon is element 14 with 4 valent electrons and sulfur is element 16 with 6 valence electrons. Thus, sulfur has stronger electronegativity (tendency to attract bonded electrons towards itself). This means that sulfur will hold oxygen tighter to itself so the hydrogen bond to it can be more easily separated from it. $HCH_{3}SO_{2}$ is more acidic in aqueous solution.

b) In $H_{3}PO_{4}$ , phosphorus holds one double bond with oxygen and three OH group equally. To show an acidic tendency, phosphorus would need to let go one hydrogen out of one of OH groups. In $H_{3}PO_{3}$ , phosporus holds two double bong with oxygen, one OH and one hydrogen, all single and lonely, ready to leave phosphorus and show acidic characteristics in aqueous solution. Thus, $H_{3}PO_{3}$ is more acidic compound.

C) In all Cl acids, the electron density is placed around Cl so the more oxygen around Cl, the more acidic will be the chemical. This is comparable to an oxidation state - the bigger oxidation state, the stronger acid will be:

$HClO_{4} ^{+7} >HClO_{3}^{+5} >HClO_{2}^{+3} >HClO_{}^{+1}$

$HClO_{2}$ can reasonably be expected to be more acidic in aqueous solution.

4 0

2 years ago