Question

Use this reaction: 3NaOH + H3PO4--> Na3PO4 + 3H20
How many mL of .225 M NaOH will react with 4.568g H3PO4?
The answer for this is 621 mL
b) What is the maximum amount of Na3PO4 that is formed from the reaction of 25.00 mL of .1050 M NaOH and 15.00 mL of .08650 M H3PO4? answer is .141g Na3PO4

Idk how to do the steps in dimensional analysis. Please help!

Answer 1

a)
Find the moles of H3PO4 we will use this formula to do just that
moles = grams/molecular weight

H3PO4 has a molecular weight of 97.99g/mol
thus
4.568g/97.99g/mol = 0.0466mol
0.04662mol
Now we know the moles of H3PO4 we are reacting the NaOH with, we now have to figure out how many moles of NaOH that can react with.

To do this we look at the reaction formula, notice that for every 3 parts of NaOH we have 1 part of H3PO4

so we just multiply 0.04662mol * 3 = 0.13986moles
so now we know that 0.13986 moles of NaOH will react with the amount of H3PO4 we have
Now just use the formula
Molarity = moles/Litres

We know the molarity of the NaOH solution and we know the amount of moles we need so solve for the volume and you have your answer
0.255M = 0.13986moles/xLitres
0.13986moles/0.255M = 0.548L or 548mL

I think your answer may be incorrect I double checked my math and I'm still not getting 621mL

b)
Just follow the method above and you should be able to solve this, use the limiting reagent to calculate though as it will dictate how much product is formed. The limiting reagent being the reactant that is used up the fastest and in the lowest concentration.

The limiting agent in this reaction is the H3PO4.

Answer 2

a)
n1=moles of NaOH
n2=moles of H3PO4
as @Australopithecus said it mast be
n1=3n2
C1*V1=3*(m2/MM2)
.......
V1=3*m2/MM2*C1
V1=3*4.568/98*0.255
V1=0.548L

for you to find 621mL you mast do a mistake and that mistake is to punch 0.225 instead 0.255 :):):)
so either given 0.255 is wrong
either answer 0.621 is wrong