The combustion of glucose is represented by the following balanced equation: C6H12O6+6 O2→6 H2O+6 CO2. The reaction uses 1 gram of both C6H12O6 and O2. What is the percent yield if 0.45 g of H2O is produced? a 0.558% b 100% c 0.31% d 80%

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04-03-2020
Chemistry

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The combustion of glucose is represented by the following balanced equation: C6H12O6+6 O2→6 H2O+6 CO2. The reaction uses 1 gram of both C6H12O6 and O2. What is the percent yield if 0.45 g of H2O is produced? a 0.558% b 100% c 0.31% d 80%

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Alleei Alleei · Answer 1 · 2020-03-05T06:39:29+01:00

Answer : The correct option is, (d) 80 %

Solution : Given,

Mass of [tex]C_6H_{12}O_6[/tex] = 1 g

Mass of [tex]O_2[/tex] = 1 g

Molar mass of [tex]C_6H_{12}O_6[/tex] = 180 g/mole

Molar mass of [tex]O_2[/tex] = 32 g/mole

Molar mass of [tex]H_2O[/tex] = 18 g/mole

First we have to calculate the moles of [tex]C_6H_{12}O_6[/tex] and [tex]O_2[/tex].

[tex]\text{ Moles of }C_6H_{12}O_6=\frac{\text{ Mass of }C_6H_{12}O_6}{\text{ Molar mass of }C_6H_{12}O_6}=\frac{1g}{180g/mole}=0.00555moles[/tex]

[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{1g}{32g/mole}=0.0312moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]C_6H_{12}O_6+6O_2\rightarrow 6H_2O+6CO_2[/tex]

From the balanced reaction we conclude that

As, 6 mole of [tex]O_2[/tex] react with 1 mole of [tex]C_6H_{12}O_6[/tex]

So, 0.0312 moles of [tex]O_2[/tex] react with [tex]\frac{0.0312}{6}=0.0052[/tex] moles of [tex]C_6H_{12}O_6[/tex]

From this we conclude that, [tex]C_6H_{12}O_6[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]H_2O[/tex]

From the reaction, we conclude that

As, 6 mole of [tex]O_2[/tex] react to give 6 mole of [tex]H_2O[/tex]

So, 0.0312 mole of [tex]O_2[/tex] react to give 0.0312 mole of [tex]H_2O[/tex]

Now we have to calculate the mass of [tex]H_2O[/tex]

[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]

[tex]\text{ Mass of }H_2O=(0.0312moles)\times (18g/mole)=0.562g[/tex]

Theoretical yield of [tex]H_2O[/tex] = 0.562 g

Experimental yield of [tex]H_2O[/tex] = 0.45 g

Now we have to calculate the percent yield of the reaction.

[tex]\% \text{ yield of reaction}=\frac{\text{ Experimental yield of }H_2O}{\text{ Theoretical yield of }H_2O}\times 100[/tex]

[tex]\% \text{ yield of reaction}=\frac{0.45g}{0.562g}\times 100=80\%[/tex]

Therefore, the percent yield of reaction is, 80 %

darren201 darren201 · Answer 2 · 2020-05-21T21:59:19+02:00

darren201 darren201

21-05-2020

Answer:

The yield would D. 80%!

Explanation:

Since 1 gram of O2 only produces 0.56 g of H2O, whereas 1 g of C6H12O6 produces 0.60 g of H2O, the O2 is the limiting reagent.

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