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3 years ago

Which compound could serve as a reactant in a neutralization reaction?(1) NaCl (3) CH3OH(2) KOH (4) CH3CHO

Chemistry

2 answers:

NikAS [45]3 years ago

8 0

KOH could serve as a reactant in a neutralization reaction

Serjik [45]3 years ago

5 0

Answer: (2) $KOH$

Explanation: When an acid reacts with the base then it undergoes neutralization to form a salt and water.

An acid is a substance that ionizes in the water to give hydrogen ion and have pH less than 7.

$HX\rightarrow H^++X^-$

A base is a substance that ionizes in the water to give hydroxide ion and has pH more than 7.

$MOH\rightarrow M^++OH^-$

Thus $KOH$ being a base can be a reactant in neutralization reaction.

$NaCl$ being a salt can be a product in a neutralization reaction.

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A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

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Answer: The percent yield is, 93.4%

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First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

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