Question

Question 1(Multiple Choice Worth 4 points)

Answer 1

Answer:

0.60 L.

Explanation:

How many moles of $\rm H_2SO_4$?

$n(\mathrm{H_2SO_4}) = c\cdot V = 4.0\times 0.30 = \rm 1.2\; mol$.

How many moles of $\rm NaOH$ will react with all that 1.2 moles of $\rm H_2SO_4$?

Balance the equation:

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\;H_2O$.

The coefficient in front of $\rm H_2SO_4$ is 1. The coefficient in front of $\rm NaOH$ is 2. Hence the ratio:

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$.

Therefore

$\displaystyle n(\mathrm{NaOH}) = \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} \cdot n(\mathrm{H_2SO_4}) = 2\times 1.2 = \rm 2.4\;mol$.

What will be the volume of the $\rm NaOH$ solution?

$\displaystyle V(\mathrm{NaOH}) = \frac{n}{c} = \frac{2.4}{4.0} = \rm 0.60\;L$.