Question

To neutralize the acid in 10.0 mL of 18.0 M H2SO4 that was accidentally spilled on a laboratory bench top, solid sodium bicarbonate was used. The container of sodium

bicarbonate was known to weigh 155.0 g before this use and out of curiosity its mass was measured as 144.5 g afterwards. The reaction that neutralizes sulfuric acid this way is as follows: H2SO4 + 2 NaHCO3 ---> Na2SO4 + 2 CO2 + 2 H2O

Was sufficient sodium bicarbonate used? Calculate the limiting reactant and the maximum yield in grams of sodium sulphate.

Answer 1

Step-by-step explanation:

The first step is to determine the amount of H2SO4 that was spilled:

moles of H2SO4 = concentration x volume

= 18.0 mol/L x 0.0100 L

= 0.180 mol

Next, we need to determine which reactant is limiting. To do this, we can compare the amount of moles of H2SO4 with the amount of moles of NaHCO3 that would be needed to completely neutralize it. From the balanced equation, we know that 1 mole of H2SO4 reacts with 2 moles of NaHCO3:

moles of NaHCO3 needed = 2 x moles of H2SO4

= 2 x 0.180 mol

= 0.360 mol

The mass of sodium bicarbonate used is the difference between the initial mass and the final mass:

mass of NaHCO3 used = 155.0 g - 144.5 g

= 10.5 g

We can convert the mass of NaHCO3 used to moles:

moles of NaHCO3 used = mass / molar mass

= 10.5 g / 84.01 g/mol

= 0.125 mol

Since we have more moles of H2SO4 than moles of NaHCO3 needed, the limiting reactant is NaHCO3. This means that H2SO4 is in excess.

To determine the maximum yield of Na2SO4, we need to calculate the theoretical yield based on the limiting reactant:

moles of Na2SO4 = 1/2 x moles of NaHCO3 used

= 1/2 x 0.125 mol

= 0.0625 mol

mass of Na2SO4 = moles x molar mass

= 0.0625 mol x 142.04 g/mol

= 8.88 g

Therefore, the maximum yield of Na2SO4 that can be obtained is 8.88 g. Since the actual amount of NaHCO3 used is less than what is needed to completely react with the H2SO4, we can conclude that not enough sodium bicarbonate was used.