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Double displacement reaction occurs when the negatively and the positively charged ions of the compounds interchange their position when the new compound in the product is formed.

<h3>Which is a double displacement reaction?</h3>

In the <u>first reaction</u>, the element lead (Pb) is not an ion and no exchange of the cations or the anions occurs hence it is not a double displacement reaction.

In the <u>second reaction</u>, one compound is getting split into two new compounds and is a decomposition reaction.

In the <u>third reaction</u>, $\rm Br^{-}$ in $\rm 2NaBr$ and $\rm (OH)_{2}$ in $\rm Ca(OH)_{2}$ are anions and cations and exchanges their places to produce two new compounds $\rm CaBr_{2} + 2NaOH.$

In the <u>fourth reaction</u>, a single compound is generated as the result of the combination reaction and hence, it is not the double displacement reaction.

Therefore, $\rm 2NaBr + Ca(OH)_{2} \rightarrow CaBr_{2} + 2NaOH$ is the double displacement reaction.

Learn more about double displacement reaction here:

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3 0

2 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

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Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$

$\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g$

Therefore, the theoretical yield of HCl is, 56.1735 grams

7 0

3 years ago