Final answer:
To find the volume of 3.00 M H₂SO₄ needed, calculate the mass of Ba(NO₃)₂ in the sample based on its weight percent, convert this mass to moles, and use the molar ratio from the balanced equation to determine the moles and hence the volume of H₂SO₄ required.
Step-by-step explanation:
To determine how many milliliters of 3.00 M H₂SO₄ are required to react with a solid containing 23.2 wt% Ba(NO₃)₂, we first calculate the mass of Ba(NO₃)₂ in the solid. This is done by multiplying the total mass of the solid (4.35 g) by the weight percent of Ba(NO₃)₂ (23.2%), which gives us the mass of Ba(NO₃)₂ that needs to react.
Next, we convert the mass of Ba(NO₃)₂ to moles using its molar mass. From the balanced equation Ba²⁺ + SO⁴²⁺ → BaSO₄(s), we see that the reaction occurs in a 1:1 molar ratio between Ba(NO₃)₂ and H₂SO₄. This allows us to calculate the number of moles of H₂SO₄ needed, which can then be converted to volume (in milliliters) using the concentration (3.00 M) of the H₂SO₄ solution.
First, we need to calculate the moles of Ba(NO3)₂ in 4.35 g using its molar mass. The molar mass of Ba(NO3)₂ is 261.34 g/mol, so the moles of Ba(NO3)₂ is:
moles of Ba(NO3)₂ = mass of Ba(NO3)₂ / molar mass of Ba(NO3)₂
moles of Ba(NO3)₂ = 4.35 g / 261.34 g/mol = 0.01664 mol
According to the balanced equation, the mole ratio between Ba(NO3)₂ and H₂SO₄ is 1:1. So, the moles of H₂SO₄ needed is also 0.01664 mol.
Now we can calculate the volume of 3.00 M H₂SO₄ needed using its molarity:
volume of H₂SO₄ = moles of H₂SO₄ / molarity of H₂SO₄
volume of H₂SO₄ = 0.01664 mol / 3.00 mol/L = 0.00555 L = 5.55 mL