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.

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    $\begingroup$ 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 $\ce{NaHSO4}$.

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

| improve this answer | |

Mol of $\ce{NaOH}$= $3/40$

Mol of $\ce{H2SO4} = 4.9/98.1 = 49/981$

If you compare the mol of both of them, you will see that the mol of $\ce{NaOH}$ is in excess. You would now do $\ce{NaOH}$ - $\ce{H2SO4}$. By doing this you have used up all of the $\ce{H2SO4}$.

$3/40 - 49/981= 983/39240$

This is the mol of $\ce{Na2SO4}$. You can clearly see that the mol of $\ce{NaOH}$ is in excess.

Due to the molar ratio of $\ce{H2SO4}$= $\ce{Na2SO4}$ (1=1). The number of mol of $\ce{Na2SO4}$ made should be the amount of $\ce{H2SO4}$ which has all been used up.

Therefore, $49/981 = 0.0499 $

$0.0499 * 142.1 = 7.01 g $

I think answer given by OP is wrong.


More information given below.

Here is the equation with some information. enter image description here

To make this explanation as easy as possible, the initial moles are only to 3 significant figures(So the actual answer at the end will not be accurate) and have been multiplied by 1000.

So you can see that $\ce{NaOH} $ makes up $75 mol $ and $\ce{H2SO4}$ is $55 mol $ . The "left in excess row " is the $\ce{2NaOH}$ -$\ce{H2SO4}$. You can easily see with theses numbers that there will be 20 mol of $\ce{2NaOH}$ left over and all of the $55 mol $ of $\ce{H2SO4}$ will be used up.

if $55 mol$ of $\ce{H2SO4}$ has been used up then 55 mol of $\ce{Na2SO4}$ has been produced, due to $1 to 1 $ molar ratio of those two molecules. Then divide by 1000 as this was simply so the numbers are easier to look at.

| improve this answer | |
  • $\begingroup$ Wouldn't you need 2 mol of $\ce{NaOH}$ to get one mol of $\ce{Na2SO4}$? $\endgroup$ – Gyro Gearloose Dec 28 '15 at 15:40
  • $\begingroup$ Oh, I see, you use all $\ce{H2SO4}$ to produce $\ce{NaHSO4}$, then you have 983/39240 mols of $\ce{NaOH}$ left to procude exactely that much mol $\ce{Na2SO4}$. $\endgroup$ – Gyro Gearloose Dec 28 '15 at 15:56
  • $\begingroup$ @GyroGearloose Have a read and let me know how it is. $\endgroup$ – Viv Dec 28 '15 at 18:26
  • $\begingroup$ To interleave, the OPs calculation would be correct if $\ce{Na2SO4}$ would be completely insoluble and participate completely? $\endgroup$ – Gyro Gearloose Dec 28 '15 at 18:47
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    $\begingroup$ yes, I see that. It looks like we agree on every detail but still can't close that matter. I propose to sit back and wait for the original poster to give some input. $\endgroup$ – Gyro Gearloose Dec 28 '15 at 20:18

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