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2 years ago

Write the formula unit equation for this reaction occuring in water:

Chemistry

1 answer:

larisa [96]2 years ago

5 0

The formula unit equation for the given reaction occurring in water is

CaS(aq) + FeBr₂(s) → FeS(s) + CaBr₂(aq).

<h3>What is chemical reaction?</h3>

Chemical reactions are those reaction in which reactant molecules react with each other for the formation of product.

When calcium sulfide reacts with iron (II) bromide for the formation of iron(II) sulfide and calcium bromide will be represented through the following equation:

CaS(aq) + FeBr₂(s) → FeS(s) + CaBr₂(aq)

Above reaction best explains the formula unit of given reaction as all valency of atoms are satisfied there.

Hence, the required formula unit is CaS(aq) + FeBr₂(s) → FeS(s) + CaBr₂(aq).

To know more about chemical reactions, visit the below link:

brainly.com/question/26018275

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<h3>What is the chemistry of living systems called?</h3>

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2 years ago

How to find the number of neutrons in an isotope?

Alexeev081 [22]

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4 0

3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

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

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

4 years ago

How would you prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0% by mass with a density of 1.19 g/mL

hoa [83]

<h2>Step 1 : Identify the given </h2>

Volume = 250mL

Density = 1.19 g/ML

<h2>Step 2 . Calculate the mass of HCL </h2>

Density = mass/volume

∴Mass = Density * Volume

= 1.19g/mL* 250mL

= 297,5g

<h2>Step 3 : Calculate the total mass of the solution, given that concentration HCL is 38% </h2>

Mass of the total solution can be calculated by the following :

38% = Mc /297.5 * 100

Mc = 38/100 *297.5

= 113.05grams

• Finally, this means that mass of the total solution of 0.125M HCL i,s 113grams, ,you would use this mass to prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0%

6 0

1 year ago