Determining the mass of sodium sulfate that forms when reacting sulfuric acid with sodium hydroxide

Question

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.

Question:

$3\ \mathrm{g}$ 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.

Answer 1

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 $\ce{NaHSO4}$.

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

Answer 2

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.

Update:

More information given below.

Here is the equation with some information.

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.