What is the minimum volume of 10.0 M H2SO4 necessary to react completely with 16.2g of Na2CrO4?

To determine the minimum volume of 10.0 M H2SO4 necessary to react completely with 16.2g of Na2CrO4, you need to use the stoichiometry of the balanced chemical equation. Here's how you can get the answer:

Step 1: Write the balanced chemical equation:
2 Na2CrO4 + H2SO4 -> Na2SO4 + Cr2(SO4)3 + H2O

Step 2: Calculate the molar mass of Na2CrO4:
2 × (22.99 g/mol) + (51.99 g/mol) + 4 × (16.00 g/mol) = 261.97 g/mol

Step 3: Convert the mass of Na2CrO4 to moles:
16.2 g ÷ 261.97 g/mol = 0.0619 mol Na2CrO4

Step 4: Use the stoichiometry of the balanced equation to find the moles of H2SO4:
From the balanced equation, 2 moles of Na2CrO4 react with 1 mole of H2SO4.
Thus, 0.0619 mol Na2CrO4 × (1 mol H2SO4 / 2 mol Na2CrO4) = 0.0309 mol H2SO4

Step 5: Calculate the minimum volume of 10.0 M H2SO4:
Molarity (M) = moles of solute / volume of solution (in liters)
0.0309 mol H2SO4 = (10.0 mol/L) × (volume of H2SO4 in liters)
Volume of H2SO4 = 0.0309 mol H2SO4 / 10.0 mol/L = 0.00309 L = 3.09 mL

Therefore, the minimum volume of 10.0 M H2SO4 necessary to react completely with 16.2g of Na2CrO4 is 3.09 mL.