I need to know what volume of liquid phosphoric acid at 85%, H3PO4 with density of 1.685 will replace 5 grams of powdered NaH2PO4. H2O.
thanks
This question makes no sense. What are you trying to replace? No amount H3PO4 will replace the Na, for example, but you could replace the PO4 with an eqivalent amount of H3PO4.
ok, I am try to get an equal amount of phosphorous.
Currently using 6 mls of a 75% liquid phosporic acid for an original equation that uses 5g of powdered phosphoric acid. My new solution is 85% and I wanted to check calculation.
To determine the volume of liquid phosphoric acid needed to replace 5 grams of powdered NaH2PO4·H2O, we need to use stoichiometry and the density of the phosphoric acid.
First, let's write the balanced chemical equation for the reaction between NaH2PO4·H2O and H3PO4:
NaH2PO4·H2O + H3PO4 -> NaH2PO4 + H2O
From the equation, we can see that 1 mole of NaH2PO4·H2O reacts with 1 mole of H3PO4.
To find the number of moles of NaH2PO4·H2O, we need to convert the mass of NaH2PO4·H2O to moles using its molar mass.
The molar mass of NaH2PO4·H2O is:
Na (22.99 g/mol) + H2 (2.02 g/mol) + P (30.97 g/mol) + O4 (16.00 g/mol) + H2O (18.02 g/mol) = 156.01 g/mol.
Using the equation:
moles = mass / molar mass
moles of NaH2PO4·H2O = 5 g / 156.01 g/mol = 0.0320 moles
Since the reaction is 1:1, we need 0.0320 moles of H3PO4.
Now, let's calculate the volume of H3PO4 using its density of 1.685 g/mL.
To find the volume, we can use the equation:
volume = mass / density
mass of H3PO4 = moles of H3PO4 * molar mass of H3PO4
mass of H3PO4 = 0.0320 moles * (3*(1.01 g/mol) + 1*(16.00 g/mol)) = 2.58 g
volume = 2.58 g / 1.685 g/mL = 1.53 mL
Therefore, to replace 5 grams of NaH2PO4·H2O, you will need approximately 1.53 mL of liquid phosphoric acid at 85% concentration and with a density of 1.685 g/mL.