Question

Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to you there is a small jar labeled "15.00 g sodium bicarbonate". Will this be enough sodium bicarbonate to neutralize the spilled sulfuric acid? Show your work and state all reasoning!

Answer 1

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

   $H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)$

Next we will calculate how many moles of $H_2SO_4$ are present in 85.00 mL of 1.500 M sulfuric acid.

As,       Molarity = $\frac{\text{moles of solute}}{\text{liters of solution }}$

            1.500 M = $\frac{n}{0.08500 L }$

                    n = 0.1275 mol $H_2SO_4$

Now set up and solve a stoichiometric conversion from moles of $H_2SO_4$  to grams of $NaHCO_3$. As, the molar mass of $NaHCO_3$ is 84.01 g/mol.

 $0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})$

                 = 21.42 g $NaHCO_3$

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.