Question

A 6g sample of carbon allowed to burn in 20g of oxygen gas produce carbon dioxide. After the reaction, the mass of unreacted oxygen is 4 g. What mass of carbon dioxide was produced? show with steps

Answer 1

Answer:

Approximately $22\;\rm g$.

Explanation:

Look up the relative atomic mass data of carbon and oxygen on a modern periodic table:

• $\rm C$: $12.011$.
• $\rm O$: $15.999$.

Calculate the formula mass of $\rm O_2$ and $\rm CO_2$:

$M(\rm O_2) = 2 \times 15.999 = 31.998\; \rm g \cdot mol^{-1}$.

$M(\rm CO_2) = 12.011 + 2 \times 15.999 = 44.009 \; \rm g \cdot mol^{-1}$.

The question suggests that $20\; \rm g - 4\; \rm g = 16\; \rm g$ of $\rm O_2$ took part in this reaction. Calculate the number of moles of molecules in that $16\; \rm g$ of $\rm O_2\,$:

$\displaystyle n(\mathrm{O_2}) = \frac{m(\mathrm{O_2})}{M(\mathrm{O_2})} = \frac{16\; \rm g}{31.998\; \rm g \cdot mol^{-1}} \approx 0.500\; \rm mol$.

Since $\rm O_2$ is in excess, it would react with $\rm C$ at a one-to-one ratio to produce $\rm CO_2$:

$\rm C + O_2 \to CO_2$.

Notice the ratio between the coefficient of $\rm O_2\!$ and $\rm CO_2$: $\displaystyle\frac{n(\mathrm{CO_2})}{n(\mathrm{O_2})} = \frac{1}{1}$. For each mole of $\rm O_2$ that is consumed, one mole of $\rm CO_2\!$ will be produced.

The question implies that $16\; \rm g$, which is approximately $0.500\; \rm mol$, of $\rm O_2$ was consumed in this reaction. Accordingly, $0.500\; \rm mol\!$ of $\rm CO_2$ will be produced.

Calculate the mass of that $0.500\; \rm mol\!$ of $\rm CO_2$:

$m(\mathrm{CO_2}) = n(\mathrm{CO_2})\cdot M(\mathrm{CO_2}) \approx 22\; \rm g$.