Determine the volume of 2.3 M H3PO4 that can completely react with?
340 mLs of a 5.1 M solution of NaOH according to the reaction:
3 NaOH + H3PO4 --> Na3PO4 + 3 H2O
stuck on a homework problem..pls help!
you want three moles of NaOH for each mole of H3PO4
moles NaOH= .340*5.1
moles H3PO4= 1/3 * .340*5.1
volumeof H3PO4= molesH3PO4/2.3 in liters.
Sure, I can help you with that! To determine the volume of 2.3 M H3PO4 that can completely react with 340 mL of a 5.1 M solution of NaOH, you'll need to use the concept of stoichiometry and the given balanced chemical equation.
The balanced chemical equation for the reaction is:
3 NaOH + H3PO4 -> Na3PO4 + 3 H2O
From the balanced equation, we can see that the stoichiometric ratio between NaOH and H3PO4 is 3:1. This means that for every 3 moles of NaOH, we need 1 mole of H3PO4 to react completely.
To find the number of moles of NaOH in the 340 mL (0.340 L) solution, we can use the formula:
moles = volume (L) x concentration (M)
moles of NaOH = 0.340 L x 5.1 M = 1.734 moles
Since the stoichiometric ratio between NaOH and H3PO4 is 3:1, we can say that 1.734 moles of NaOH will react with 1/3 x 1.734 moles of H3PO4.
moles of H3PO4 = 1/3 x 1.734 moles = 0.578 moles
Now, we can use the relationship between moles and concentration to find the volume of the 2.3 M H3PO4 solution required to react with 0.578 moles of H3PO4.
volume = moles / concentration
volume of H3PO4 = 0.578 moles / 2.3 M = 0.251 L = 251 mL
Therefore, the volume of the 2.3 M H3PO4 solution that can completely react with 340 mL of the 5.1 M NaOH solution is 251 mL.