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

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:

According to the reaction:

of glucose on combustion gives of is partial

of the glucose on combustion  given ......... grams of

Therefore the final mass is "0.1652 g".

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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

ch3 [ch2]14cooh

explanation:

fire could start, buildings collapsing, and ash falling from the sky.

explanation:

this is because the buildings or plants could catch on fire and the buildings would most likely collapse. lastly ash would fall because all the burning would cause ash to fall from the sky.

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