Find the volume of 0.250M sulfuric acid necessary to react completely with 79.3g sodium hydroxide
To find the volume of 0.250M sulfuric acid necessary to react completely with 79.3g of sodium hydroxide, you need to follow the following steps:
Step 1: Determine the molar mass of sodium hydroxide (NaOH)
The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol. Hydrogen (H) has a molar mass of 1.01 g/mol. Therefore, the molar mass of NaOH is:
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Total: 39.99 g/mol
Step 2: Calculate the number of moles of sodium hydroxide (NaOH)
To calculate the number of moles, divide the given mass (79.3 g) by the molar mass of NaOH:
79.3 g / 39.99 g/mol = 1.98 moles of NaOH
Step 3: Determine the balanced chemical equation
The balanced equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
Step 4: Determine the stoichiometric ratio
From the balanced equation, the stoichiometric ratio between sulfuric acid and sodium hydroxide is 1:2. This means that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.
Step 5: Calculate the number of moles of sulfuric acid (H2SO4)
Since the stoichiometric ratio is 1:2, the number of moles of sulfuric acid is half the number of moles of sodium hydroxide:
1.98 moles of NaOH / 2 = 0.99 moles of H2SO4
Step 6: Calculate the volume of sulfuric acid
The given concentration of sulfuric acid is 0.250M, which means that there are 0.250 moles of H2SO4 in every liter of solution.
To calculate the volume of sulfuric acid, divide the number of moles (0.99 moles) by the molarity (0.250M):
Volume = moles / molarity
Volume = 0.99 moles / 0.250M
Volume = 3.96 liters
Therefore, the volume of 0.250M sulfuric acid necessary to react completely with 79.3g of sodium hydroxide is 3.96 liters.
To find the volume of sulfuric acid necessary to react completely with sodium hydroxide, we need to use the balanced chemical equation and stoichiometry.
The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equation, we can see that one mole of sulfuric acid reacts with 2 moles of sodium hydroxide. Therefore, we need to find the number of moles of sodium hydroxide to determine the number of moles of sulfuric acid.
First, let's calculate the number of moles of sodium hydroxide using its molar mass.
Molar mass of NaOH:
Na (sodium) = 23.0 g/mol
O (oxygen) = 16.0 g/mol
H (hydrogen) = 1.0 g/mol
Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g/mol
Now, we can calculate the number of moles of sodium hydroxide:
Number of moles = mass / molar mass
Number of moles = 79.3 g / 40.0 g/mol
Next, we can determine the number of moles of sulfuric acid needed to react with the calculated moles of sodium hydroxide. Since the stoichiometric ratio is 1:2, the number of moles of sulfuric acid will be twice the number of moles of sodium hydroxide.
Number of moles of sulfuric acid = 2 * Number of moles of sodium hydroxide
Finally, we can calculate the volume of sulfuric acid using the molarity (0.250M = 0.250 moles per liter) and the number of moles of sulfuric acid:
Volume (L) of sulfuric acid = Number of moles of sulfuric acid / Molarity
This calculation will give us the volume of 0.250M sulfuric acid necessary to react completely with 79.3g sodium hydroxide.
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
mols NaOH = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols NaOH to mols H2SO4.
Now convert mols H2SO4 to volume.
M H2SO4 = mols H2SO4/L H2SO4. You know mols and M, solve for L.