Answer:
0.60 L.
Explanation:
How many moles of [tex]\rm H_2SO_4[/tex]?
[tex]n(\mathrm{H_2SO_4}) = c\cdot V = 4.0\times 0.30 = \rm 1.2\; mol[/tex].
How many moles of [tex]\rm NaOH[/tex] will react with all that 1.2 moles of [tex]\rm H_2SO_4[/tex]?
Balance the equation:
[tex]\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\;H_2O[/tex].
The coefficient in front of [tex]\rm H_2SO_4[/tex] is 1. The coefficient in front of [tex]\rm NaOH[/tex] is 2. Hence the ratio:
[tex]\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2[/tex].
Therefore
[tex]\displaystyle n(\mathrm{NaOH}) = \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} \cdot n(\mathrm{H_2SO_4}) = 2\times 1.2 = \rm 2.4\;mol[/tex].
What will be the volume of the [tex]\rm NaOH[/tex] solution?
[tex]\displaystyle V(\mathrm{NaOH}) = \frac{n}{c} = \frac{2.4}{4.0} = \rm 0.60\;L[/tex].
