Question

In the combustion analysis of 0.1127 g of glucose (c6h12o6), what mass, in grams, of co2 would be produced?

Answer 1

The combustion produces 0.1652 g CO₂.

Balanced equation = C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

Moles of C₆H₁₂O₆ = 0.1127 g C₆H₁₂O₆ × (1 mol C₆H₁₂O₆/180.16 g C₆H₁₂O₆)

= 0.000 6256 mol C₆H₁₂O₆

Moles of CO₂ = 0.000 6256 mol C₆H₁₂O₆× (6 mol CO₂/1 mol C₆H₁₂O₆)

= 0.003 753 mol CO₂

Mass of CO₂ = 0.003 753 mol CO₂ × (44.01g CO₂/1 mol CO₂) = 0.1652 g CO₂

Answer 2

Combustion analysis is the basic analytical method of natural solids and fluids. The quantity of carbon-hydrogen, oxygen, and even nitrogen oxides in the substances could be determined. Joseph Louis Gay-Lussac had also invented that method, and the calculation can be defined as follows:

$\bold{C_{6}H_{12}O_{6} + 6O_{2} \longrightarrow 6CO_{2}+6H_{2}O}$

According to the reaction:

$\bold{180\ g}$ of glucose on combustion gives $\bold{ 264\ g}$ of $\bold{CO_2}$ is partial

$\bold{0.1127\ g }$ of the glucose on combustion  given ......... grams of $\bold{CO_2}$

               $\bold{=\frac{0.1127 \times 264}{180}}\\\\\bold{=\frac{29.7528}{180}}\\\\\bold{=0.1652}$

Therefore the final $\bold{CO_2}$ mass is "0.1652 g".

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