in a lab there is a spill of 47.5ml of .500 mol/l NaOH how many grams of Sulfuric acid is needed to neutralizethe spill
You need to know how many moles of H2SO4 per liter.
You can get at the grams another way without knowing the concn of H2SO4. You need to know concn IF we wanted to know the volume of H2SO4 needed; however, we only need to know grams. That is done as follows:
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
mols NaOH = M x L = ?
Use the coefficients to convert mols NaOH to mols H2SO4. Therefore,
mols H2SO4 = 1/2 x mols NaOH
Then mols H2SO4 = grams H2SO4/molar mass H2SO4. You know molar mass and you know mols, substitute and solve for grams H2SO4.
Sorry. I did not read carefully enough.
To calculate the amount of sulfuric acid needed to neutralize the spill of NaOH, we need to use the balanced chemical equation that represents the neutralization reaction between NaOH and H2SO4.
The balanced equation for the reaction is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
From the balanced equation, we can see that the stoichiometric ratio between NaOH and H2SO4 is 2:1. This means that for every 2 moles of NaOH, we need 1 mole of H2SO4 to react completely.
First, let's calculate the number of moles of NaOH in the spill:
moles of NaOH = concentration (mol/L) × volume (L)
= 0.500 mol/L × 0.0475 L
= 0.02375 mol
Since the molar ratio between NaOH and H2SO4 is 2:1, we need half the number of moles of H2SO4 to react completely with the NaOH.
Now, let's calculate the number of moles of H2SO4 needed:
moles of H2SO4 = (moles of NaOH) / 2
= 0.02375 mol / 2
= 0.011875 mol
Finally, we can calculate the mass of sulfuric acid (H2SO4) required using its molar mass.
The molar mass of H2SO4 is:
2(1.01 g/mol for hydrogen) + 32.07 g/mol (for sulfur) + 4(16.00 g/mol for oxygen)
= 98.09 g/mol
mass of H2SO4 = moles of H2SO4 × molar mass of H2SO4
= 0.011875 mol × 98.09 g/mol
= 1.16 g
Therefore, approximately 1.16 grams of sulfuric acid (H2SO4) is required to neutralize the spill of NaOH.