How many milliliters of 0.10 M H3PO4 could be neutralized by 20.0g of NaOH?
H3PO4(aq)+3NaOH(aq)->3H2O(l)+NaPO4(aq)
To what volume must 30ml of a 2.5m NaOH solution be dillutedto make a 0.4m solution?
What amount of H3PO4 (in moles) should react with these 20.0g of NaOH?
To determine the number of milliliters of 0.10 M H3PO4 that could be neutralized by 20.0g of NaOH, we need to use stoichiometry.
First, we need to find the number of moles of NaOH. To do this, we divide the mass of NaOH by its molar mass. The molar mass of NaOH is 22.99 g/mol (Na: 22.99 g/mol, O: 16.00 g/mol, H: 1.01 g/mol).
Number of moles of NaOH = 20.0g NaOH / 22.99 g/mol = 0.870 mol NaOH
According to the balanced chemical equation, we know that 1 mole of H3PO4 reacts with 3 moles of NaOH. Therefore, the number of moles of H3PO4 is three times the number of moles of NaOH.
Number of moles of H3PO4 = 3 * 0.870 mol NaOH = 2.610 mol H3PO4
Now, we can use the molarity of H3PO4 and the number of moles of H3PO4 to calculate the volume of 0.10 M H3PO4.
Molarity (M) = moles / volume (L)
We rearrange the formula to solve for volume:
Volume (L) = moles / Molarity
Volume of 0.10 M H3PO4 = 2.610 mol H3PO4 / 0.10 mol/L = 26.1 L
Finally, to convert liters to milliliters, we multiply the volume by 1000:
Volume of 0.10 M H3PO4 = 26.1 L * 1000 = 26,100 mL
Therefore, 26,100 milliliters of 0.10 M H3PO4 could be neutralized by 20.0 grams of NaOH.