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3 years ago

Calculate the moles of 12.354 grams of c6h12o6

Chemistry

1 answer:

gavmur [86]3 years ago

5 0

The correct answer is 0.06857 moles.

C₆H₁₂O₆, that is, glucose has six carbons, twelve hydrogens, and six oxygen atoms. The atomic weight of C, H and O are as follows:

Six atoms of carbon = 6 × 12.01 g = 72.06 g

Twelve atoms of hydrogen = 12 × 1.008 g = 12.096 g

Six atoms of oxygen = 6 × 16.00 g = 96.00 g

So, the molar mass of C₆H₁₂O₆ is 72.06 g + 12.096 g + 96.0 g = 180.156 g.

It can also be written in the form as 180.16 g of C₆H₁₂O₆ is equal to 1 mole of C₆H₁₂O₆or 180.16 g/mole (as the molar mass)

Now, there is a need to find moles of 12.354 grams of C₆H₁₂O₆. So, the final conversion is:

12.354 g C₆H₁₂O₆ × 1 mole of C₆H₁₂O₆ / 180.16 g C₆H₁₂O₆

= 0.06857 moles

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rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o

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Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

Initial important note:

Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.

1) Air:

Source: internet

Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.

2) Exhaled air:

Source: internet.

Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂

3) Air from the decomposition of H₂O₂.

In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:

i) Chemical Equation:

H₂O₂ (g) → H₂ (g) + O₂ (g)

ii) mole ratio of the products 1 mol H₂ : 1 mol O₂

iii) convert moles into mass (grams)

1 mol H₂ × 2 × 1.008 g/mol = 2.016 g

1 mol O₂ × 2 × 15.999 g/mol = 31.998 g

Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%

4) Air from the decomposition of NaHCO₃:

i) chemical equation:

2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)

ii) mole ratio: take into account only the gases in the products:

1 mol CO₂ (g) : 1 mol H₂O

iii) mass in grams

CO₂: molar mass ia approximately 44.01 g/mol

H₂O: molar mass is approximately 18.02 g/mol

iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.

5) The result is:

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