In order to neutralize the acid in 10.0 mL of 18.0 mol/L sulfuric acid that was spilled on a laboratory bench, sodium bicarbonate (baking soda) was used. The container of baking soda had a mass of 155.00g before this use, and out of curiosity it's mass was measured as 144.50g afterward.
Write a balanced equation for the reaction between sulfuric acid and sodium bicarbonate.
was sufficient sodium bicarbonate used? identify by calculation the limiting and excess reactants.
Oh, let me sprinkle some humor into this chemistry question! 🎪
Alrighty, here's the balanced equation for the reaction between sulfuric acid and sodium bicarbonate:
H2SO4 + 2NaHCO3 --> Na2SO4 + 2H2O + 2CO2
Now, onto the fun part! To determine if sufficient sodium bicarbonate was used, we need to calculate the moles of sulfuric acid and sodium bicarbonate:
Moles of sulfuric acid: 10.0 mL * 0.0180 mol/L = 0.18 mol
Moles of sodium bicarbonate: (155.00 g - 144.50 g) / 84.01 g/mol = 0.124 mol
Since we have fewer moles of sodium bicarbonate (0.124 mol) than sulfuric acid (0.18 mol), the sodium bicarbonate is the limiting reactant. It's like when you have too little popcorn at the movies and run out before the movie ends - bummer!
So, there you have it! The amount of sodium bicarbonate used was not sufficient to fully neutralize the sulfuric acid. You might need to bring in some reinforcements or break out the clown car for more baking soda next time. 🤡💨
The balanced equation for the reaction between sulfuric acid (H2SO4) and sodium bicarbonate (NaHCO3) is:
2 NaHCO3 + H2SO4 -> Na2SO4 + 2 H2O + 2 CO2
To determine if sufficient sodium bicarbonate was used, we need to calculate the moles of sulfuric acid and sodium bicarbonate.
First, let's calculate the moles of sulfuric acid:
Moles of H2SO4 = concentration of H2SO4 (in mol/L) * volume of H2SO4 (in L)
= 18.0 mol/L * 0.0100 L
= 0.180 mol
Next, let's calculate the moles of sodium bicarbonate:
Moles of NaHCO3 = mass of NaHCO3 (in g) / molar mass of NaHCO3
= (155.00 g - 144.50 g) / 84.01 g/mol
= 10.50 g / 84.01 g/mol
= 0.125 mol
According to the balanced equation, the molar ratio between sulfuric acid and sodium bicarbonate is 1:2. This means that 1 mole of H2SO4 reacts with 2 moles of NaHCO3.
Since we have 0.180 mol of H2SO4 and 0.125 mol of NaHCO3, the limiting reactant is NaHCO3 because it is present in lesser quantity. This implies that all of the NaHCO3 would be consumed in the reaction, and some H2SO4 may be left over.
Therefore, there is not sufficient sodium bicarbonate to completely neutralize the acid.
To write a balanced equation for the reaction between sulfuric acid and sodium bicarbonate, we first need to understand the chemical formulas of the two compounds. The chemical formula for sulfuric acid is H2SO4, and the chemical formula for sodium bicarbonate is NaHCO3.
The balanced equation for the reaction between sulfuric acid and sodium bicarbonate is as follows:
H2SO4 + 2NaHCO3 → Na2SO4 + 2H2O + 2CO2
Now, let's determine if sufficient sodium bicarbonate was used by calculating the limiting and excess reactants.
Step 1: Find the number of moles of sulfuric acid (H2SO4) spilled.
Given:
Volume of sulfuric acid = 10.0 mL
Molarity of sulfuric acid = 18.0 mol/L
First, convert the volume of sulfuric acid from milliliters (mL) to liters (L):
10.0 mL ÷ 1000 = 0.010 L
Now, calculate the number of moles of sulfuric acid:
Number of moles = Molarity × Volume
Number of moles = 18.0 mol/L × 0.010 L = 0.18 mol
Step 2: Find the number of moles of sodium bicarbonate (NaHCO3) used.
Given:
Mass of sodium bicarbonate before = 155.00 g
Mass of sodium bicarbonate after = 144.50 g
First, calculate the change in mass of sodium bicarbonate:
Change in mass = Mass before - Mass after
Change in mass = 155.00 g - 144.50 g = 10.50 g
Next, convert the mass of sodium bicarbonate to moles using its molar mass:
Molar mass of NaHCO3 = (22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 3(16.00 g/mol)) = 84.01 g/mol
Number of moles = Mass / Molar mass
Number of moles = 10.50 g / 84.01 g/mol ≈ 0.125 mol
Step 3: Determine the limiting and excess reactants.
The balanced equation provides a stoichiometric ratio between sulfuric acid and sodium bicarbonate as 1:2. From the above calculations, we have determined that 0.18 mol of sulfuric acid was used and 0.125 mol of sodium bicarbonate was used.
Since the stoichiometric ratio of sulfuric acid to sodium bicarbonate is 1:2, we can compare the number of moles used for each reactant.
Sulfuric acid: 0.18 mol
Sodium bicarbonate: 0.125 mol (×2 to match the stoichiometric ratio = 0.250 mol)
Comparing 0.18 mol and 0.250 mol, we can see that the sulfuric acid is the limiting reactant, and the sodium bicarbonate is the excess reactant.
In conclusion, more than sufficient sodium bicarbonate was used, and sulfuric acid is the limiting reactant.
H2SO4 + 2NaHCO3 ==> Na2SO4 + 2H2O + 2CO2
mass NaHCO3 = 155.00-144.50 = 10.50 g used.
mols NaHCO3 = grams/molar mass = about 0.125
If all were to be used it would produce how many mols H2O. That is 0.125 mols NaHCO3 x (2 mols H2O/2 mols NaHCO3) = 0.125 mols CO2.
mols H2SO4 = M x L = 18.0 x 0.010 = 0.18 mols.
If all were to be used it would produce how many mols H2O? That's
0.010 mols H2SO4 x (2 mols H2O/1 mol H2SO4) = 0.010 x 2/1 = 0.020
In limiting reagent problems, the SMALLER number is the correct value; therefore, 0.02 mols CO2 will be produced, the H2SO4 is the limiting reagent, and NaHCO3 is the excess reagent. Therefore, all of the H2SO4 will be neutralized and NaHCO3 will be left over unreacted.