Question

What is the mass of the precipitate formed when 50mL of 16.9% solution of $AgN{O_3}$ is mixed with 50mL of 5.8% W/v NaCl solution?
(A) 7g
(B) 14g
(C) 28g
(D) 3.5g

Answer 1

Hint: To solve this question we should know to calculate the number of moles. It is important to be aware of the reaction between $AgN{O_3}$ and Nacl as we have to find the mass of precipitate formed, so we should know what is the precipitate formed to solve this question.

Complete step by step answer:
Firstly, we will write the reaction between $AgN{O_3}$ and NaCl
$AgN{O_3}$ reacts with NaCl to give AgCl and $NaN{O_3}$. The yellow precipitate of AgCl is obtained.
\[AgN{O_3} + NaCl \to AgCl + NaN{O_3}\]……………..equation 1
Molecular mass of $AgN{O_3}$= 169
Molecular mass of NaCl = 58.5
Number of moles = $\dfrac{{Mass(g)}}{{Molecular{\text{ mass}}}}$
-Let's calculate the number of moles $AgN{O_3}$:
Number of moles = $\dfrac{{Mass(g)}}{{Molecular{\text{ mass}}}}$
Number of moles of $AgN{O_3}$= $\dfrac{{50 \times \dfrac{{6.9}}{{100}}}}{{169}}$
\[ = \dfrac{{50 \times 6.9}}{{100 \times 169}}\]
= 0.05mole
-Let's calculate the number of moles of NaCl:

Number of moles = $\dfrac{{Mass(g)}}{{Molecular{\text{ mass}}}}$
Number of moles of NaCl $ = \dfrac{{50 \times \dfrac{{5.8}}{{100}}}}{{58.5}}$
 \[ = \dfrac{{50 \times 5.8}}{{100 \times 58.5}}\]
 = 0.05 mole
With reference to equation 1, we can calculate the mass of AgCl:
By applying POAC on Ag
Number of moles Ag in $AgN{O_3}$= Number of moles Ag in AgCl
Number of moles of AgCl = 0.05mole…(ii)
Molecular mass of AgCl = 143.5………..(iii)
Mass of AgCl = Number of moles $ \times $ Molecular mass of AgCl ………(iv)
Substitute (ii) and (iii) in (iv)
= 0.05 $ \times $ 143.5
= 7.16g

Thus, option A is the correct answer.

Note: The things to remember to solve these types of questions.
-The formula to calculate number of moles:
Number of moles = $\dfrac{{Mass(g)}}{{Molecular{\text{ mass}}}}$
-Remember the chemical equation 1 as it is widely used.
\[AgN{O_3} + NaCl \to AgCl + NaN{O_3}\]
 The yellow precipitate of AgCl is obtained. so we need to find the mass of AgCl .