Question

Use this reaction: 3NaOH + H3PO4--> Na3PO4 + 3H20
what is the maximum amount of Na3PO4 that is formed from the reaction of 25ml of .1020M NaOH and 15ml of .08650 M H3PO4? The answer is given to me for checking, it is 0.141g Na3PO4.
I am just starting at stoichiometry problems, so I need all the help i can get! thank you!

Answer 1

We're given starting values for both reactants, so we first need to figure out which is the limiting reagent. To do this, we need to find how many moles of each reactant there are, and divide that value by their stoichiometric coefficient (the number beside them in the chemical equation above). We divide by the stoichiometric coefficient to take into account the fact that a different number of molecules for each reactant is required for the reaction to proceed.

For NaOH: moles = 0.025 L x 0.1020 M = 2.55 x 10^(-3) mol
2.55 x 10^(-3) mol / 3 = 8.5 x 10^(-4)
For H3PO4: moles = 0.015 L x 0.08650 M = 1.30 x 10^(-3) mol
1.30 x 10^(-3) mol / 1 = 1.30 x 10^(-3)

NaOH has the lower number, so it is our limiting reagent. This means NaOH will be used up before H3PO4, and so will determine how much Na3PO4 is produced. From our chemical equation, the ratio of Na3PO4 formed to NaOH used is 1:3.

1 Na3PO4 / 3 NaOH = x Na3PO4 / 2.55 x 10^(-3) mol NaOH
x = 8.5 x 10^(-4) mol

This is how many moles of Na3PO4 are produced. We can convert it to grams by multiplying it by the molar mass of Na3PO4, which is 164 g/mol: 8.5 x 10^(-4) mol x 164 g/mol = 0.1394 g.

My answer is a bit different from yours because my rounding was probably a bit different, but that's how you do these problems! Hope that helps!

Answer 2

Thank you! this really helped! :)