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3 years ago

How many grams are in 1.946 moles of nacl

Chemistry

1 answer:

Ilia_Sergeevich [38]3 years ago

7 0

Answer:

113.8g

Explanation:

Statement of problem: mass of 1.946mole of NaCl

Given parameters:

Number of moles of NaCl = 1.946mole

Unknown: mass of NaCl

Solution

To find the mass of NaCl, we apply the concept of moles which expresses the relationship between number of moles and mass according to the equation below:

Number of moles = $\frac{mass}{molar mass}$

To find the molar mass of NaCl:

the atomic mass of Na = 23g

atomic mass of Cl = 35.5g

Molar mass of NaCl = (23 + 35.5) = 58.5gmol⁻¹

Mass of NaCl = Number of moles x molar mass of NaCl

Mass of NaCl = 1.946 x 58.5 = 113.8g

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Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 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:

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

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

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

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By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

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Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

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