Question

Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O. Suppose 62. g of sulfuric acid is mixed with 33.8 g of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

Approximately
`21\; \rm g`.

Step-by-step explanation:

`\rm H_2SO_4` (a diprotic acid) reacts with
`\rm NaOH` (a monoprotic base) at a one-to-two ratio:

`\rm 2\; NaOH\, (s) + H_2SO_4\, (aq) \to Na_2SO_4\; (aq) + 2\; H_2O\, (l)`.

In other words, if
`n(\mathrm{NaOH})` and
`n(\mathrm{H_2SO_4})` represent the number of moles of the two compounds reacted, then:

`\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = (1)/(2)`.

Look up the relative atomic mass data on a modern periodic table:

• 
  `\rm H`:
  `1.008`.
• 
  `\rm S`:
  `32.06`.
• 
  `\rm O`:
  `15.999`.
• 
  `\rm Na`:
  `22.990`.

Calculate the (molar) formula mass of
`\rm H_2SO_4` and
`\rm NaOH`:

`M(\mathrm{H_2SO_4}) = 2 * 1.008 + 32.06 + 4 * 15.999 = 98.072\; \rm g \cdot mol^(-1)`.

`M(\mathrm{NaOH}) = 22.990 + 15.999 + 1.008 = 39.997\; \rm g \cdot mol^(-1)`.

Calculate the number of moles of formula units in that
`33.8\; \rm g` of
`\rm NaOH`:

`\begin{aligned}n(\mathrm{NaOH}) &= \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} \\ &= (33.8\; \rm g)/(39.997\; \rm g \cdot mol^(-1)) \approx 0.845\; \rm mol\end{aligned}`.

Apply the ratio
`\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = (1)/(2)` to find the (maximum) number of moles of
`\rm H_2SO_4` that would react with the
`33.8\; \rm g` of
`\rm NaOH`:

`\begin{aligned}n(\mathrm{H_2SO_4}) &= \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} \cdot n(\mathrm{NaOH})\\ &= (1)/(2) * 0.845 \approx 0.4225\; \rm mol\end{aligned}`.

Calculate the mass of that
`0.4225\; \rm mol` of
`\rm H_2SO_4`:

`\begin{aligned}m(\mathrm{H_2SO_4}) &= n(\mathrm{H_2SO_4}) \cdot M(\mathrm{H_2SO_4})\\ &= 0.4225 \; \rm mol * 98.072\; \rm g \cdot mol^(-1) \approx 41.435\; \rm g \end{aligned}`.

When the maximum amount of
`\rm H_2SO_4` is reacted, the minimum would be in excess. Hence, the minimum mass of

`62\; \rm g - 41.435\; \rm g \approx 21\; \rm g` (rounded to two significant figures.)