what mass NaOH is required to react exactly with 25.0 ml of 3.0 M H2S04
2NaOH + H2SO4 ==> 2H2O + Na2SO4
How many mols H2SO4 do you have? That is M x L = ?
That will take twice that for mols NaOH needed. Convert to grams by g = mols x molar mass.
6 grams
6g
To calculate the mass of NaOH required to react with a given volume and concentration of H2SO4, you can use the concept of stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between NaOH (sodium hydroxide) and H2SO4 (sulfuric acid) is as follows:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
To get the mass of NaOH required, you need to follow these steps:
Step 1: Convert the given volume of H2SO4 to moles.
- Volume of H2SO4 = 25.0 mL = 0.0250 L
- Concentration of H2SO4 = 3.0 M
- Moles of H2SO4 = Volume × Concentration = 0.0250 L × 3.0 M = 0.075 mol
Step 2: Use the stoichiometry of the balanced equation to determine the moles of NaOH needed to react with the moles of H2SO4.
- From the balanced equation, the stoichiometric ratio between NaOH and H2SO4 is 2:1.
- Moles of NaOH = 0.075 mol H2SO4 × (2 mol NaOH / 1 mol H2SO4) = 0.150 mol NaOH
Step 3: Convert the moles of NaOH to grams using the molar mass of NaOH.
- The molar mass of NaOH = 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.99 g/mol
- Mass of NaOH = 0.150 mol NaOH × 39.99 g/mol = 5.999 g (rounded to three decimal places)
Therefore, approximately 6.0 grams of NaOH is required to react exactly with 25.0 mL of 3.0 M H2SO4.