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

Approximately
`22\;\rm g`.

Step-by-step 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 * 15.999 = 31.998\; \rm g \cdot mol^(-1)`.

`M(\rm CO_2) = 12.011 + 2 * 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})} = (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})} = (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`.