I have come across simple problem in the chemistry book which I cannot solve. The question is below. I would appreciate if you have any solution to this.


$3\ \mathrm{g}$ of $\ce{NaOH}$ are mixed with $4.9\ \mathrm{g}$ of $\ce{H2SO4}$. How much sodium sulfate $\ce{Na2SO4}$ will form?

(The answer is $3.55\ \mathrm{g}$)

Note: I have calculated it as $5.33\ \mathrm{g}$ of $\ce{Na2SO4}$ don’t know where I am mistaken. It doesn’t seem to be a typo in the question because I couldn’t solve other similar problems in other books wither. There is something I always miss.

  • 2
    $\begingroup$ You seem to be assuming that all $\ce{Na}$ would go to $\ce{Na2SO4}$, and the unused $\ce{H2SO4}$ (which is in excess) would stay that way, i.e. as $\ce{H2SO4}$. This is not quite so. $\endgroup$ – Ivan Neretin Dec 28 '15 at 8:01

Rather than working with funny fractions, let’s use decimals.

$$m(\ce{H2SO4}) = 4.9~\mathrm{g}\\ m(\ce{NaOH}) = 3~\mathrm{g}\\ M(\ce{H2SO4}) = 98.09~\mathrm{g \cdot mol^{-1}}\\ M(\ce{NaOH}) = 40.00~\mathrm{g \cdot mol^{-1}}\\ M(\ce{Na2SO4}) = 142.05~\mathrm{g \cdot mol^{-1}}\\ ~% \\ n(\ce{H2SO4}) = 50~\mathrm{mmol}\\ n(\ce{NaOH}) = 75~\mathrm{mmol}$$

Remember when doing stoichiometry to correctly consider your reaction equations. In this case: $$\ce{H2SO4 + 2NaOH -> Na2SO4 + 2 H2O}$$

If we want to do this, we see that we don’t have enough $\ce{NaOH}$ — it is present only in substoichiometric amounts. We would need $100~\mathrm{mmol}~\ce{NaOH}$ for a full reaction with $\ce{H2SO4}$:

$$\frac{n(\ce{NaOH})}{n(\ce{H2SO4})} = \frac{2}{1}\\ n(\ce{NaOH}) = 2 n (\ce{H2SO4})\\ n(\ce{NaOH}) = 2 \times 50~\mathrm{mmol}= 100~\mathrm{mmol}$$

If we now assumed that all $\ce{NaOH}$ would participate in the formation of $\ce{Na2SO4}$, we would arrive at $37.5~\mathrm{mmol}~\ce{Na2SO4}$ or $5.4~\mathrm{g}\ \ce{Na2SO4}$. But that would leave free $\ce{H2SO4}$ lying around. Instead, consider a stepwise process:

$$\ce{H2SO4 + NaOH -> NaHSO4 + H2O}\\ \ce{NaHSO4 + NaOH -> Na2SO4 + H2O}$$

We see that we need to use the first $50~\mathrm{mmol}\ \ce{NaOH}$ to generate $50~\mathrm{mmol}\ \ce{NaHSO4}$; with $25~\mathrm{mmol}\ \ce{NaOH}$ remaining. These $25~\mathrm{mmol}\ \ce{NaOH}$ can react with the $\ce{NaHSO4}$ to generate $25~\mathrm{mmol}\ \ce{Na2SO4}$ and leave $25~\mathrm{mmol}$ unreacted $\ce{NaHSO4}$.

$$n(\ce{Na2SO4}) = 25~\mathrm{mmol}\\ m(\ce{Na2SO4}) = n \cdot M = 25~\mathrm{mmol} \times 142.05~\mathrm{g \cdot mol^{-1}} = 3.55~\mathrm{g}$$


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