2 KOH(l) + H3PO4(l) --> K2HPO4(l) + 2 H2O(l)

a) How many mL of 0.450M KOH is needed to react with 60.0mL of 0.250M H3PO4?

a) 0.060L x 0.250M = 0.015 mol H3PO4
0.015mol H3PO4 x 2mol KOH / 1mol H3PO4 = 0.03 mol KOH
0.03 mol KOH x 1L/0.450 mol = 0.06667L = 66.67mL

ANSWER IN SIG FIG 67 ML

b) How many mL of H2O is produced? (** Don’t forget to convert mL to L and back**)

0.03 mol H2O x (18 g/mol) = 0.54 g H2O x 1 mL / 1 g = 0.54 mL H2O.

ANSWER IN SIG FIG 54 ML H20

66.67 is right and you are allowed 3 s.f. (60.0, 0.450, 0.250 all have 3 s.f. so you round to 66.7

etc. I think you need to restudy s.f.

In support of DrBob's suggestion, see my note on sig.figs. in previous posted problem.

It's been my experience that significant figures are the one element of laboratory reports that students seem to misunderstand & misuse more than anything else. There is 'practical' importance in understanding 'why' significant figures are needed. Not intending to be rude, but science, for the most part, is a 'laboratory exercise', not a theoretical calculation. The devices we use for data collection should always be kept in mind when calculating and reporting numerical results. The general rule that governs such application is => the results should reflect the measuring device with the least degree of accuracy in the experimental study. This gives credibility and practical significance to reported results. All the best, Doc :-)
Doc :-)

In support of DrBob's suggestion, see my note on sig.figs. in previous posted problem.

It's been my experience that significant figures are the one element of laboratory reports that students seem to misunderstand & misuse more than anything else. There is 'practical' importance in understanding 'why' significant figures are needed. Not intending to be rude, but science, for the most part, is a 'laboratory exercise', not a theoretical calculation. Students understandably seem to loose this significance in the process of working problems from a theoretical perspective. The devices we use for data collection should always be kept in mind when calculating and reporting numerical results. The general rule that governs such application is => the results should reflect the measuring device with the least degree of accuracy in the experimental study. This gives credibility and practical significance to reported results. All the best, Doc :-)
Doc :-)

Oops, didn't mean to post twice.

To answer part a of the question, you need to use the given equation balanced with stoichiometric coefficients:

2 KOH(l) + H3PO4(l) -> K2HPO4(l) + 2 H2O(l)

First, calculate the moles of H3PO4 using the given concentration and volume:

0.060 L x 0.250 M = 0.015 mol H3PO4

Next, use the stoichiometric coefficients to find the moles of KOH needed to react with the moles of H3PO4:

0.015 mol H3PO4 x (2 mol KOH / 1 mol H3PO4) = 0.03 mol KOH

Finally, convert the moles of KOH to volume using the given concentration:

0.03 mol KOH x (1 L / 0.450 mol) = 0.06667 L = 66.67 mL

The answer, rounded to the correct significant figures, is 67 mL.

Moving on to part b of the question, we need to determine the volume of H2O produced. Since the stoichiometric coefficient for H2O is 2, we know that for every 2 moles of KOH reacted, 2 moles of H2O are produced.

Given that 0.03 moles of KOH reacted, we can calculate the moles of H2O produced:

0.03 mol KOH x (2 mol H2O / 2 mol KOH) = 0.03 mol H2O

Next, convert the moles of H2O to mass using the molar mass of H2O (18 g/mol):

0.03 mol H2O x (18 g/mol) = 0.54 g H2O

Finally, convert the mass of H2O to volume using the density of water (1 g/mL):

0.54 g H2O x (1 mL / 1 g) = 0.54 mL H2O

The answer, rounded to the correct significant figures, is 54 mL of H2O.