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3 years ago

What volume of 0.117 M HCl is needed to neutralize 28.67ml of 0.137 m KOH?

1 answer:

3 0

To determine the volume of the acid needed in the reaction, we first have to know the reaction involved in the process. Since it is a neutralization reaction, then it should produce a salt and water. The reaction would be:

HCl + KOH = KCl + H2O

We use this reaction to relate the substances. We calculate as follows:

0.137 M KOH (.02867 L solution ) = 3.92779x10^-3 mol KOH
3.92779x10^-3 mol KOH ( 1 mol HCl / 1 mol KOH ) = 3.92779x10^-3 mol HCl

Volume of HCl = 3.92779x10^-3 mol HCl / 0.117 M HCl
= 0.03357 L HCl or 33.57 mL HCl needed

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3 years ago

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3 years ago

Read 2 more answers

During a laboratory experiment, you discover that an enzyme-catalyzed reaction has a ∆G of -20 kcal/mol. If you double the amoun

Dennis_Churaev [7]

Answer:

dG will be the same -20 kcal/mol

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3 years ago

What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O

nadya68 [22]

Answer:

$V=43.46mL$

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

$3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O$

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

$n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4$

Then, given the molarity, it is possible to obtain the milliliters as follows:

$V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL$

Best regards!

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3 years ago