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Eddi Din [679]
2 years ago
5

Calculate the moles of 12.354 grams of c6h12o6

Chemistry
1 answer:
gavmur [86]2 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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<u>Answer:</u> Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

<u>Explanation:</u>

To calculate the number of moles, we use the equation

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}   ....(1)

  • <u>For ozone:</u>

Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol

  • <u>For nitric oxide:</u>

Given mass of nitric oxide = 0.635 g

Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol

For the given chemical equation:

O_3+NO\rightarrow O_2+NO_2

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = \frac{1}{1}\times 0.0172=0.0172moles of nitric oxide

As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ozone is considered as a limiting reagent because it limits the formation of product.

  • Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = \frac{1}{1}\times 0.0172=0.0172moles of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

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