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2 years ago

Calculate the molality and van’t Hoff factor (i) for the following aqueous solutions:

1 answer:

5 0

The formula for molality is m = moles of solute / kilograms of solvent. In problem solving involving molality, we sometimes need to use additional formulas to get to the final answer. One formula we need to be aware of is the formula for density, which is d = m / v, where d is density, m is mass and v is volume.

<h3>What molality means?</h3>

Molality (m), or molal concentration, is the amount of a substance dissolved in a certain mass of solvent. It is defined as the moles of a solute per kilograms of a solvent.

<h3>What is molality vs molarity?</h3>

Learn how molarity and molality differ! The molality of a solution is equal to the moles of solute divided by the mass of solvent in kilograms, while the molarity of a solution is equal to the moles of solute divided by the volume of solution in liters.

Learn more about molality here:

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Calcium oxide CaO = 40+16 = 56
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56/14 = 4
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3 years ago

In the following chemical reaction between H_2 and Cl_2 to produce HCl, what is the mass of HCl produced and leftover reactants

ira [324]

<u>Answer:</u> The total amount of leftover reactants and HCl is 12.79 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For hydrogen gas:</u>

Given mass of hydrogen gas = 0.36 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{0.36g}{2g/mol}=0.18mol$

<u>For chlorine gas:</u>

Given mass of chlorine gas = 12.41 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

$\text{Moles of chlorine gas}=\frac{12.41g}{71g/mol}=0.175mol$

The chemical equation for the reaction of hydrogen gas and chlorine gas is:

$H_2+Cl_2\rightarrow 2HCl$

By Stoichiometry of the reaction:

1 moles of chlorine gas reacts with 1 mole of hydrogen gas

So, 0.175 moles of chlorine gas will react with = $\frac{1}{1}\times 0.175=0.175mol$ of hydrogen gas

As, given amount of hydrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, chlorine gas is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (hydrogen gas) = [0.18 - 0.175] = 0.005 moles

By Stoichiometry of the reaction

1 moles of chlorine gas produces 2 moles of HCl

So, 0.175 moles of chlorine gas will produce = $\frac{2}{1}\times 0.175=0.350$ moles of HCl

Now, calculating the mass of hydrogen gas left and HCl from equation 1, we get:

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

Moles of excess hydrogen gas = 0.005 moles

Putting values in equation 1, we get:

$0.005mol=\frac{\text{Mass of excess hydrogen gas}}{2g/mol}\\\\\text{Mass of excess hydrogen gas}=(0.005mol\times 2g/mol)=0.01g$

<u>For HCl:</u>

Molar mass of HCl = 36.5 g/mol

Moles of HCl = 0.350 moles

Putting values in equation 1, we get:

$0.350mol=\frac{\text{Mass of HCl}}{36.5g/mol}\\\\\text{Mass of HCl}=(0.350mol\times 36.5g/mol)=12.78g$

Total mass of HCl and leftover reactants = [12.78 + 0.01] = 12.79 g

Hence, the total amount of leftover reactants and HCl is 12.79 grams

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4 years ago