Question

500 ml of 0.25M $N{{a}_{2}}S{{O}_{4}}$ solution is added to an aqueous solution of 15g of $BaC{{l}_{2}}$ resulting in the formation of a white precipitate of insoluble $BaS{{O}_{4}}$ . How many moles and how many grams of $BaS{{O}_{4}}$ are formed?

Answer 1

We will first find the moles of $N{{a}_{2}}S{{O}_{4}}$ by the formula :
$Molarity=\dfrac{number\text{ }of\text{ }moles}{volume\text{ }of\text{ }solution}$
and moles of $BaC{{l}_{2}}$. Then we will find, mass of $BaS{{O}_{4}}$ which is equal to:
$Mass=Number\text{ }of\text{ }moles\times molecular\text{ }mass$

Complete Solution :
- We have taken $N{{a}_{2}}S{{O}_{4}}$ and added it to a solution of $BaC{{l}_{2}}$ . And as a result of which we got a white precipitate of $BaS{{O}_{4}}$, and NaCl.
- We can write the reaction as:
$N{{a}_{2}}S{{O}_{4}}+BaC{{l}_{2}}\to BaS{{O}_{4}}+2NaCl$

-Now, we will see the number of moles of $N{{a}_{2}}S{{O}_{4}}$, we are being provided with molarity=0.25.
And the volume of solution is 500 ml. So, we will find number of moles by the formula: $Molarity=\dfrac{number\text{ }of\text{ }moles}{volume\text{ }of\text{ }solution}$
$0.25=\dfrac{number\text{ }of\text{ }moles}{500}\times 1000$
Here, we have converted ml into litres that’s why we multiplied it by 1000.
So, we can write it as:

& Moles=\left( \dfrac{0.25}{2} \right) \\\ & =0.125 \\\ \end{aligned}$$ So, we can say that the moles of $N{{a}_{2}}S{{O}_{4}}$=0.125 \- Now, for $BaC{{l}_{2}}$ the molecular mass = $\begin{aligned} & 131+2\times 35.5 \\\ & =131+71 \\\ & =202gm \\\ \end{aligned}$ - So, moles of $BaC{{l}_{2}}$ is equal to: $$\begin{aligned} & \dfrac{given\text{ }mass}{molecular\text{ }mass} \\\ & =\dfrac{15}{202} \\\ & =0.07425moles \\\ \end{aligned}$$ \- Now, by comparing the moles of and we get to know that the moles of $BaC{{l}_{2}}$ is less. So, $BaC{{l}_{2}}$ is having 0.074 moles , therefore moles of $BaS{{O}_{4}}$ formed is also equal to 0.0074 moles. \- Now we will find, mass of $BaS{{O}_{4}}$ which is equal to: $$Mass=Number\text{ }of\text{ }moles\times molecular\text{ }mass$$ $$\begin{aligned} & =0.074\times \left( 137+32+64 \right) \\\ & =17.242gm \\\ \end{aligned}$$ \- Hence, we can conclude that 0.0074 moles and 17.242 grams of $BaS{{O}_{4}}$ are formed. **Note:** \- We should not forget to convert the unit of volume given in millilitre into litres. \- We should not get confused in terms of normality and molarity. Normality is the number equivalent of solute dissolved per litre of solution. And, Molarity is defined as the number of moles of solute per litre of solution. The unit of normality is N. The unit of molarity is M.