how many milliliters of 0.125 M H2SO4 are needed to neutralize 0.200 g of NaOH?
To find the number of milliliters of 0.125 M H2SO4 needed to neutralize 0.200 g of NaOH, we need to use the concept of stoichiometry.
1. Start by finding the moles of NaOH:
- The molar mass of NaOH (sodium hydroxide) is 22.99 g/mol + 16.00 g/mol + 1.008 g/mol = 39.998 g/mol.
- Divide the given mass (0.200 g) by the molar mass to get the number of moles:
0.200 g NaOH / 39.998 g/mol = 0.00500 mol NaOH.
2. The balanced equation for the neutralization reaction between H2SO4 and NaOH is:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O.
3. Based on the balanced equation, we see that 2 moles of NaOH react with 1 mole of H2SO4. Therefore, 0.00500 mol NaOH will react with half that amount of H2SO4.
- Moles of H2SO4 needed = 0.00500 mol NaOH / 2 = 0.00250 mol H2SO4.
4. Now, we need to find the volume of the 0.125 M H2SO4 solution that would contain this amount of H2SO4.
- The volume of the solution can be calculated using the formula:
Volume (in liters) = Moles / Molarity.
Volume (in liters) = 0.00250 mol H2SO4 / 0.125 mol/L = 0.02000 L.
5. Convert the volume to milliliters by multiplying by 1000:
Volume (in milliliters) = 0.02000 L × 1000 mL/L = 20.00 mL.
Therefore, 20.00 milliliters of 0.125 M H2SO4 are needed to neutralize 0.200 g of NaOH.
To determine the number of milliliters of 0.125 M H2SO4 needed to neutralize 0.200 g of NaOH, we need to use the concept of stoichiometry.
First, we need to write the balanced chemical equation for the reaction between H2SO4 and NaOH:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. Next, we need to calculate the number of moles of NaOH in 0.200 g.
The molar mass of NaOH is:
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Total: 22.99 + 16.00 + 1.01 = 40.00 g/mol
Using the formula: moles = mass / molar mass, we can calculate the number of moles of NaOH:
moles = 0.200 g / 40.00 g/mol = 0.005 moles
Since the balanced equation shows that 1 mole of H2SO4 reacts with 2 moles of NaOH, we can set up a proportion:
0.005 moles NaOH / 2 moles NaOH = x moles H2SO4 / 0.125 moles H2SO4
Simplifying the proportion, we find:
x moles H2SO4 = (0.005 moles NaOH / 2 moles NaOH) * 0.125 moles H2SO4
x moles H2SO4 = 0.005 * 0.125 / 2 = 0.0003125 moles H2SO4
Now, to convert moles of H2SO4 into milliliters of 0.125 M H2SO4, we need to use the formula:
moles = concentration (M) * volume (L)
In this case, we have:
0.0003125 moles H2SO4 = 0.125 M * volume (L)
To solve for volume, we rearrange the equation:
volume (L) = 0.0003125 moles H2SO4 / 0.125 M = 0.0025 L
Finally, we convert the volume from liters to milliliters:
volume (mL) = 0.0025 L * 1000 mL/L = 2.5 mL
Therefore, 2.5 milliliters of 0.125 M H2SO4 are needed to neutralize 0.200 g of NaOH.
How many grams of CO can react with .150 kg of Fe2O3?
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
mols NaOH = g/molar mass = 0.200/40 = ?
mols H2SO4 = 1/2 that. Look at the coefficients to know that.
M H2SO4 = mols H2SO4/L H2SO4. You know M and mols, solve for L and convert to mL.