Question

how many particles of CO2 can be produced from 20.0grams of O2?

Answer 1

Answer on Question #41842 – Chemistry – Inorganic Chemistry

Question:

How many particles of $\mathrm{CO}_{2}$ can be produced from 20.0 grams of $\mathrm{O}_{2}$?

Answer:

Reaction:

$$\mathrm{C} + \mathrm{O}_{2} = \mathrm{CO}_{2}$$

First of all, we will find number of moles of $\mathrm{O}_{2}$ which have been used in chemical reaction.

$$\theta = \frac{m}{M}$$

where $\theta$ – is number of moles of $\mathrm{O}_{2}$, $m$ – is mass of $\mathrm{O}_{2}$, $M$ – is molecular mass of $\mathrm{O}_{2}$.

$$M(\mathrm{O}_{2}) = 15.9949 + 15.9949 = 31.9898 \approx 32 \, \mathrm{g/mole}$$ $$\theta = \frac{20 \, \mathrm{g}}{32 \, \mathrm{g/mole}} = 0.625 \, \mathrm{mole}$$

As you see from the chemical reaction of carbon oxidation from 1 mole of $\mathrm{O}_{2}$ we can get 1 mole of $\mathrm{CO}_{2}$. And in 1 mole of substance we have $6.022 \times 10^{23}$ elementary entities or particles of the substance.

But in our case we have just 0.625 mole of $\mathrm{O}_{2}$ which can produce 0.625 mole of $\mathrm{CO}_{2}$.

In 1 mole of substance is $6.022 \times 10^{23}$ elementary particles

In 0.625 mole of substance is $X$ elementary particles

Then

$$X = \frac{0.625 \, \mathrm{mole} \times 6.022 \times 10^{23} \, \mathrm{elementary particles}}{1 \, \mathrm{mole}} = 3.764 \times 10^{23} \, \mathrm{elementary particles}$$

Answer:

$3.764 \times 10^{23}$ elementary particles of $\mathrm{CO}_{2}$ can be produced from 20.0 grams of $\mathrm{O}_{2}$.

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