[EGutierrez91]

1) 51.0 mL of 3.70 M sodium hydroxide is combined with 36.0 mL of 1.80 M magnesium chloride. What mass in grams of solid forms?

2) What is the molar concentration of a solution of sulfuric acid, H2SO4, if 8.09 mL react completely with 0.606 g of NaOH which was dissolved in 200 mL of water?

3) What is the molarity of a solution of calcium hydroxide, Ca(OH)2, if 147 mL is neutralized with 47.49 mL of a 0.276 M HCl?

4) Potassium acid phthalate, KHC8H4O4, is a crystalline solid that is available in a state of high purity, making it an excellent choice as a standard acid. What is the concentration of a NaOH solution, if it takes 55.00 mL to neutralize a 0.0396 g sample of the acid? For the neutralization reaction, KHC8H4O4(aq) --> K+(aq) + H+(aq) + C8H4O42-(aq)

I have these 4 problems left of my assignment. I honestly don't even know how to properly start them off.

1) I know the solid is Mg(OH)2, but no idea how to find the mass of it. 1 mole Mg(OH)2 is 58.32g but how do I find the number of moles?

The other ones I don't even know how to start...

any help is greatly appreciated. Thanks!

[Borek]

Start with reaction equations.

[EGutierrez91]

Start with reaction equations.

1) 2NaOH + MgCl2 --> 2NaCl + Mg(OH)2


2) H2SO4 + 2NaOH --> (Na)2SO4 + 2H20

3) Ca(OH)2 + HCl --> CaCl2 + 2H2O

4) KHC8H4O4 + NaOH --> K + H + C8H4O4 + NaOH     (not too sure about this one)

[Borek]

First one - how many moles of sodium hydroxide? How many moles of magnesium chloride? Which is a limiting reagent?

Phthalate is a weak acid. See http://www.titrations.info/acid-base-titration-solution-standardization

[Charkol]

Did you remember Avagadro's number?  6.022*10^23 pieces in one mole.

[Borek]

Did you remember Avagadro's number?  6.022*10^23 pieces in one mole.

While true, this is completely unnecessary to solve the questions.

[EGutierrez91]

First one - how many moles of sodium hydroxide? How many moles of magnesium chloride? Which is a limiting reagent?

Phthalate is a weak acid. See http://www.titrations.info/acid-base-titration-solution-standardization

NaOH = .1887mol
MgCl2 = .0648mol

Since you need two moles of NaOH for each mole of MgCl2, there is enough MgCl2 for this reaction, but the MgCl2 is still the limiting reagent.

[Borek]

So how many moles of Mg(OH)₂ will be produced?

Beware, spoon feeding is about to end, seems like you are perfectly capable of solving the question on its own, you just have to start working.

[EGutierrez91]

So how many moles of Mg(OH)₂ will be produced?

Beware, spoon feeding is about to end, seems like you are perfectly capable of solving the question on its own, you just have to start working.

The last part is where I'm actually stuck though. I'm not sure if the Mg(OH)2 is the weight of the Mg from the original one, and how do I obtain moles Mg from the MgCl2?

I know that once I have the number of moles I can get the mass. It's just obtaining the moles of Mg(OH)2.

I was thinking maybe since NaOH is .1887moles and MgCl2 is .0648moles then .0648 moles of Mg(OH)2 exist and (.0648*2) moles of NaCl (because you have 2 moles per 1 mole of Mg(OH)2 ).

Is my reasoning correct for thinking it is .0648 moles of Mg(OH)2?

[Borek]

Look at the reaction equation. Do you know how to read it?

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations

(yes, you are right about number of moles of Mg(OH)₂ being identical to number of moles of MgCl₂, but reading the page linked to won't hurt).

[EGutierrez91]

I still need help with #2 and #3. I'm not really sure how to do those.

Can you help me get started Borek?

Thanks!

[Borek]

Reaction equations please.

Take a look here: http://www.titrations.info/titration-calculation

Even if these questions don't use word "titration" they are about exactly the same procedure.