x gm og Ag was dissolved in HNO3 and the solution was treated with excess of NaCl when 2.87 g of AgCl was precipitated. find the value of x?
The molecular mass of AgCl is 143.37......
143.37g AgCl contains 108 Ag.
So , 2.87 g AgCl contains (?)
then,
(2.87×108)/143.37
= 2.16
THANKS ................
143.5 g AgCl contains Ag = 108
2.87 g Agcl contains Ag = (108×2.87)/143.5
Answer= 2.16 g
Thanks :) with regards from SHABAZ AHMED DAR (VIVEKANANDA KENDRA VIDYALAYA, BARAGOLAI, ASSAM)
To solve this problem, we need to use the concept of stoichiometry.
1. Start by writing the balanced chemical equation for the reaction:
Ag + NaCl → AgCl + NaNO3
2. Determine the molar mass of AgCl:
AgCl = Ag (108 g/mol) + Cl (35.5 g/mol)
AgCl = 143.5 g/mol
3. Calculate the number of moles of AgCl precipitated using the given mass:
Moles of AgCl = Given mass / Molar mass
Moles of AgCl = 2.87 g / 143.5 g/mol
Moles of AgCl = 0.02 mol
4. According to the balanced equation, the molar ratio of AgCl to Ag is 1:1. This means that for every 1 mole of AgCl, 1 mole of Ag is needed.
5. Therefore, the number of moles of Ag dissolved in HNO3 is also 0.02 mol.
6. Finally, calculate the mass of Ag using the number of moles and the molar mass:
Mass of Ag = Moles of Ag × Molar mass
Mass of Ag = 0.02 mol × 108 g/mol
Mass of Ag = 2.16 g
Hence, the value of x is 2.16 grams.