Question #240414

When the body consumes glucose, it requires oxygen gas. This process is also known as combustion. In a laboratory, a 0.5484 g glucose sample underwent combustion, producing carbon dioxide gas and liquid water.


a. Write the balanced chemical equation for the reaction described above.

b. How many moles of glucose are present before the reaction?

c. What is the mole relationship between glucose and carbon dioxide? Between glucose and water?

d. How many grams of CO2 is produced from the given amount of sample.


1
Expert's answer
2021-09-23T02:37:42-0400

a) C6H12O6+6O26CO2+6H2OC_6H_{12}O_6+6O_2\to 6CO_2+6H_2O

b) RMM for glucose =12×6+1×12+(16×6)=18012\times 6+1\times 12+(16\times6)=180

moles of glucose=0.5484180=0.003047moles=\frac{0.5484}{180}=0.003047moles

c) mole ratio C6H12:6CO2=1:6C_6H_{12} : 6CO_2 =1:6

mole ratio C6H12:6H2O=1:6C_6H_{12}:6H_2O=1:6

d) mole of CO2=0.003047×6=0.018282molesCO_2=0.003047\times 6=0.018282moles

RMM of CO2=12+16×2=48RMM \space of \space CO_2=12+16\times 2=48

mass of CO2=48×0.018282=0.8775gCO_2=48\times 0.018282=0.8775g


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