Calculate the amount of carbon dioxide formed when we burn 1 mole of carbon in the presence of 16 g of dioxygen. $$\ce{C + O2 -> CO2}$$
1 mole of carbon means 12 g. Therefore amount of carbon dioxide is 12 + 16 = 28. Am I correct?
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Sign up to join this communityCalculate the amount of carbon dioxide formed when we burn 1 mole of carbon in the presence of 16 g of dioxygen. $$\ce{C + O2 -> CO2}$$
1 mole of carbon means 12 g. Therefore amount of carbon dioxide is 12 + 16 = 28. Am I correct?
From the equation, $$\ce{C + O2 -> CO2},$$ we can see that one mole of $\ce{C}$ reacts with one mole of $\ce{O2}$.
Therefore, let us calculate the amount of each reactant present.
The amount of $\ce{C}$ is already given: $n(\ce{C}) = \pu{1 mol}$.
The molar mass of $\ce{O2}$ is $M(\ce{O2}) = \pu{32 g//mol}$, so $\pu{16 g}$ of $\ce{O2}$ are $\pu{0.5 mol}$.
Since one mole of $\ce{C}$ only reacts with one mole of $\ce{O2}$, there remain $\pu{0.5 mol}$ of unreacted $\ce{C}$, while $\pu{0.5 mol}$ of $\ce{C}$ react with $\pu{0.5 mol}$ of $\ce{O2}$.
In formal terms, oxygen is called the limiting reactant as it limits the amount of product formed.
From the equation, one mole of $\ce{CO2}$ is produced per each mole of $\ce{C}$. Therefore, $\pu{0.5 mol}$ of $\ce{CO2}$ is produced.
The one mole of carbon would react with one mole of dioxygen to give one mole of carbon dioxide.
$\pu{12 g}$ of carbon would therefore react with $\pu{32 g}$ of $\ce{O2}$ to give $\ce{44 g}$ of $\ce{CO2}$.
As someone else has pointed out, it would be tricky to get absolutely all the carbon to burn in oxygen unless you set it up in a particular way to do so. Probably in real life some would not burn efficiently and you might end up with carbon monoxide, which is poisonous if you breath it in. Some carbon might also stay as carbon (soot).