chem Q

[NJdian]

which one and how? thanks

What is the molarity of the salt produced in the reaction of 200 mL of 0.100 M HCl with 100 mL of 0.500 M KOH?
(a) 0.0325 M
(b) 0.0472 M
(c) 0.0667 M
(d) 0.0864 M
(e) 0.0935 M

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[mahGOL]

if the correct answer is C (0.0677M) I can explain it.

which one and how? thanks

What is the molarity of the salt produced in the reaction of 200 mL of 0.100 M HCl with 100 mL of 0.500 M KOH?
(a) 0.0325 M
(b) 0.0472 M
(c) 0.0667 M
(d) 0.0864 M
(e) 0.0935 M

 

[NJdian]

yes it is right but I dont get 0.066

 

[BirkChick]

yes it is right but I dont get 0.066

Write the Equation:
KOH + HCl -> KCl + H2O

Find the moles of KOH:
(.5 mol/1 L) * (.1 L) = .05 mol KOH

Find the moles of HCl:
(.1 mol/1 L) * (.2 L) = .02 mol HCl <--- limiting reagent

Total Volume: .2 L + .1 L = .3 L

Molarity: .02 mol / .3 L = .066667 M

 

[NJdian]

thanks
I guess I have a problem with limiting reagent
one more Q

How many grams of NaOH would be required to neutralize all the acid in 75.0 mL of 0.0900 N H2SO4?
(a) 0.540 g
(b) 0.270 g
(c) 1.32 g
(d) 0.660 g
(e) 0.859 g

 

[mahGOL]

I guess the answer is B. this is how I got it.

For acid and base to neutralize each other :

NV(of H2SO4) = NV(NaOH)
(0.09 * .075) = N'V' ( of NaOH) = mole of NaOH
mass = mole * MW = 0.09 * .075 * 40 = 0.27 g

thanks
I guess I have a problem with limiting reagent
one more Q

How many grams of NaOH would be required to neutralize all the acid in 75.0 mL of 0.0900 N H2SO4?
(a) 0.540 g
(b) 0.270 g
(c) 1.32 g
(d) 0.660 g
(e) 0.859 g

 

[NJdian]

thanks mahGol
it seems you are ready for an actual test... I am so nervous because mine is going to be in 2 weeks.....

good luck

I guess the answer is B. this is how I got it.

For acid and base to neutralize each other :

NV(of H2SO4) = NV(NaOH)
(0.09 * .075) = N'V' ( of NaOH) = mole of NaOH
mass = mole * MW = 0.09 * .075 * 40 = 0.27 g

 

[Rook]

thanks mahGol
it seems you are ready for an actual test... I am so nervous because mine is going to be in 2 weeks.....

good luck

mahGol !!!
You're a chem wiz!!!