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Licemer1 [7]
3 years ago
9

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

arbon monoxide (MW=28g/mol) with the excess O2 produces 10g of carbon dioxide? Answer in units of %
Chemistry
1 answer:
Arturiano [62]3 years ago
8 0

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

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Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of HNO_3

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M_1V_1=M_2V_2

where,

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M_2\text{ and }V_2 are the final molarity and volume of HNO_3.

We are given:

M_1=0.3403M\\V_1=160.0mL\\M_2=0.0501M\\V_2=?

Putting values in above equation, we get:

0.3403M\times 160.0mL=0.0501M\times V_2\\\\V_2=1086.79mL

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of C_6H_5NH_2 +  Volume of HNO_3

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

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Now we have to calculate the pH of the solution.

At equivalence point,

pOH=\frac{1}{2}[pK_w+pK_b+\log C]

pOH=\frac{1}{2}[14+4.87+\log (0.0437)]

pOH=8.76

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-8.76\\\\pH=5.24

Thus, the pH of the solution is, 5.24

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