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Reika [66]
3 years ago
8

21 pts. Sodium reacts with oxygen to produce sodium oxide as described by the balanced equation below. If 54.1g of sodium reacts

with excess oxygen gas to produce 61.8g of sodium oxide, what is the percent yield? Show all work. (hint: be sure to calculate theoretical yield first) 4Na + O2 --> 2Na2O
Chemistry
1 answer:
Aleksandr [31]3 years ago
8 0
First, calculate the amount fo Na2O that should be produced with the given amount of sodium using proper dimensional analysis and  the balanced chemical reaction,

   theoretical amount of Na2O = (54.1 g of Na)(1 mol Na/23 g Na)(2 mos Na2O/4 mols Na)(62 g Na2O/ 1mol Na2O)
       
      theoretical amount of Na2O = 72.9 g of Na2O


Calculation of percent yield,

                % yield = (61.8 g Na2O) / (72.9 g of Na2O) x 100%
                 % yield = 84.75%

Answer: 84.75%
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<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethyl acetate = 398 g

Molar mass of ethyl acetate = 88.11 g/mol

Putting values in above equation, we get:

\text{Moles of ethyl acetate}=\frac{398g}{88.11g/mol}=4.52mol

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{fusion}}{T}

where,  

\Delta S = Entropy change  = ?

n = moles of ethyl acetate = 4.52 moles

\Delta H_{fusion} = enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol   (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = 84.0^oC=[84+273]K=357K

Putting values in above equation, we get:

\Delta S=\frac{4.52mol\times 10500J/mol}{357K}\\\\\Delta S=132.9J/K

Hence, the entropy change of the ethyl acetate is 133. J/K

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