Question

Silver Nitrate Solution was mixed with 5.8 g of Sodium Chloride Solution to verify the law of conservation of mass 8.1 gram of Silver Chloride was formed and Sodium Nitrate formed was equal to half of the amount of Silver Nitrate Solution used. What is the amount of Silver Nitrate used and Sodium Nitrate formed? ​

Answer 1

Answer:

The amount of Silver Nitrate used is $4.6g$ and Sodium Nitrate formed is $2.3g.$ ​

Explanation:

The balanced chemical equation for the given reaction is ,

  $AgNO_{3} \ +\ NaCl$  →  $NaNO_{3}\ +\ AgCl$

Given that,

 The weight of Sodium chloride taken $=$ $5.8$ g

 The weight of Silver chloride formed $=\ 8.1$ g

The weight of Sodium Nitrate formed $=$ half of the amount of Silver Nitrate Solution taken

Let's take the weight of silver nitrate taken $=$ X g

Then,

The weight of Sodium Nitrate formed $=$ $\frac{X}{2}$ g

According to the law of conservation of mass, '' the mass of products in a chemical reaction must equal the mass of reactant"·

That is,

 The total mass of reactant $=$ The total mass of products

so,

 $X$ $+\ \ 5.8\ \ =\ \ \frac{X}{2}\ \ +\ \ 8.1$

⇒ $X\ -\ \frac{X}{2}\ -\ 2.3\ =\ 0$

⇒ $X\ -\ \frac{X}{2}\ =\ 2.3$

⇒ $\frac{X}{2}\ \ =\ \ 2.3$

⇒ $X\ =\ 4.6$

That is,

The amount of silver nitrate taken $=\ X\ =\ 4.6g$

The weight of Sodium Nitrate formed $=$ $\frac{X}{2}$ g  $=\ 2.3g$