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Kobotan [32]
3 years ago
13

In a titration, 45.0 mL of KOH is neutralized by 75.0 mL of 0.30M HBr. How much KOH is in 1.0 liter of the KOH solution ?

Chemistry
1 answer:
photoshop1234 [79]3 years ago
8 0
KOH + HBr ---> KBr + H2O

0,3 moles of HBr ---in-------1000ml
x moles of HBr-------in------75ml
x = 0,0225 moles of HBr

according to the reaction:  1 mole of KOH = 1 mole of HBr
so
0,0225 moles of HBr = 0,0225 moles of KOH

0,0225 mole of KOH------in-----45ml
x moles of KOH -----------in------1000ml
x = 0,5 moles of KOH

answer: 0,5 mol/dm³  KOH (molarity)


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The percent yield of the calcium hydroxide is 84.5%.

<h3>What is stoichiometry?</h3>

Stoichiometry enables us to obtain the mass of a substance form the equation of the reaction.

The equation of the reaction is;

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2) The reaction equation is; 2NaOH + CaCO3 --->  Ca(OH)2 + Na2CO3

Number of moles of X = 25.0 grams/100 g/mol =  0.25 moles

Number of moles of NaOH= 40/1000 L * 2 M = 0.08 moles

If 1 mole of X reacts with 2 moles of NaOH

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0.08 moles of NaOH produces 0.08 moles * 1 mole/2 moles

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Percent yield = 2.5 g/ 2.96  g * 100

= 84.5%

Learn ore about percent yield:brainly.com/question/17042787

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