Question

If 32 ml of 7.0 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?

Answer 1

The minimum mass of NaHCO3 that must be added to neutralize the spilled H2SO4 is 37.632 grams.

Step-by-step explanation:

To determine the minimum mass of NaHCO3 needed to neutralize the spilled H2SO4, we can use stoichiometry.

The balanced equation for the reaction between H2SO4 and NaHCO3 is:

H2SO4 (aq) + 2NaHCO3(aq) → Na2SO4 (aq) + 2CO2 (g) + 2H2O (l)

From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaHCO3. We are given the volume (32 mL) and concentration (7.0 M) of the H2SO4, so we can calculate the number of moles of H2SO4:

Moles H2SO4 = volume (L) x concentration (M) = 32 mL / 1000 mL/L x 7.0 M = 0.224 moles H2SO4

Now, we can use the stoichiometry of the reaction to calculate the minimum mass of NaHCO3 needed to neutralize the acid. Since the molar ratio of H2SO4 to NaHCO3 is 1:2, we can set up the following conversion:

0.224 moles H2SO4 x (2 moles NaHCO3/1 mole H2SO4) x (84.0 g NaHCO3/1 mole NaHCO3) = 37.632 g NaHCO3

Therefore, the minimum mass of NaHCO3 that must be added to the spill to neutralize the acid is 37.632 grams of NaHCO3.

Answer 2

The complete balanced reaction of this neutralization reaction is:

H2SO4 + 2NaHCO3 --> 2CO2 + 2H2O + Na2SO4

Then we calculate the moles of H2SO4 that was spilled:

moles H2SO4 = 7 mole/L * 0.032 L = 0.224 mole

From the reaction, we see that 2 moles of NaHCO3 is required for every mole of H2SO4, hence:

moles NaHCO3 = 0.224 * 2 = 0.448 mole

The molar mass of NaHCO3 is 84 g/mol. Hence the mass is:

mass NaHCO3 = 0.448 * 84

mass NaHCO3 = 37.632 grams