Determining the mass of sodium sulfate that forms when reacting sulfuric acid with sodium hydroxide
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:
$$\ce{H2SO4 + 2NaOH -&gt; Na2SO4 + 2 H2O}$$

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:

$$\ce{H2SO4 + NaOH -&gt; NaHSO4 + H2O}\\
\ce{NaHSO4 + NaOH -&gt; Na2SO4 + H2O}$$

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 $\ce{NaHSO4}$.

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