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Gelneren [198K]
2 years ago
11

How many moles of sodium sulfate will be formed if you start with 3.7 moles of sodium hydroxide?

Chemistry
1 answer:
Andrew [12]2 years ago
3 0

First check it's balanced or not

On both sides

  • Na=2
  • O=6
  • H=4
  • S=1

Its balanced

  • 2 mol NaOH produces 1 mol Na_2SO_4
  • 1 mol NaOH produces 0.5mol Na_2SO_4

3.7 mol produces=3.7(0.5)=1.850=1.8mol Na_2SO_4

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H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles
abruzzese [7]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

H_2(g)+F_2(g)\rightarrow 2HF(g)

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]

We are given:

\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = \frac{-14.1}{1}\times 2.20=-31.02J/K

Hence, the value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

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Hey!

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