What volume of a 0.50 M solution of H2SO4 is required to completely neutralize 3.0 grams of NaOH? Write the equation.
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H2SO4 + 2NaOH ==> 2H2O Na2SO4
mol NaOH = grams/molar mass
Convert mols NaOH to mols H2SO4. That will be mols H2SO4 = 1/2 mols NaOH.
M H2SO4 = mols H2SO4/L H2SO4. You know M and mols, solve for L H2SO4 and convert, if you wish, to mL.
75 mL
Why did the acid go to therapy?
Because it had some serious H2SO4 feelings!
In order to determine the volume of the H2SO4 solution required to neutralize 3.0 grams of NaOH, we first need to write the balanced chemical equation:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
The equation shows that 1 mole of H2SO4 reacts with 2 moles of NaOH, meaning there is a 1:2 ratio of H2SO4 to NaOH.
To find the moles of NaOH from the given mass, we use the formula:
moles = mass / molar mass
The molar mass of NaOH is approximately 40.00 g/mol.
moles of NaOH = 3.0 g / 40.00 g/mol = 0.075 mol
Since the ratio is 1:2, we need twice as many moles of H2SO4 as NaOH:
moles of H2SO4 = 2 * moles of NaOH = 2 * 0.075 mol = 0.15 mol
Now, we can calculate the volume of the H2SO4 solution using the formula:
volume = moles / concentration
Given that the concentration of H2SO4 is 0.50 M, we substitute the values:
volume = 0.15 mol / 0.50 mol/L = 0.30 L
Therefore, to completely neutralize 3.0 grams of NaOH, you would need a volume of 0.30 liters (or 300 mL) of the 0.50 M H2SO4 solution.
To find the volume of a solution required to neutralize a given amount of a substance, we need to use the concept of stoichiometry and the equation of the reaction.
The balanced equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. We need to convert the given mass of NaOH to moles using its molar mass.
The molar mass of NaOH is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H):
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
Now we can calculate the number of moles of NaOH:
Number of moles = Mass / Molar mass
Number of moles = 3.0 g / 40.00 g/mol = 0.075 mol
According to the stoichiometry of the reaction, 2 moles of NaOH react with 1 mole of H2SO4. Therefore, since the ratio is 2:1, 0.075 moles of NaOH will react with 0.0375 moles of H2SO4.
Now, we need to calculate the volume of the 0.50 M (molar) solution of H2SO4 required to supply 0.0375 moles of H2SO4.
Molarity (M) is defined as moles of solute divided by liters of solution:
Molarity = Moles of solute / Volume of solution (in liters)
Rearranging the formula, we can calculate the volume of the solution as:
Volume of solution (in liters) = Moles of solute / Molarity
Given that the molarity (M) is 0.50 M and the number of moles is 0.0375 mol, we can substitute these values into the formula:
Volume of solution = 0.0375 mol / 0.50 M = 0.075 L = 75 mL
Therefore, the volume of the 0.50 M solution of H2SO4 required to completely neutralize 3.0 grams of NaOH is 75 mL.