Question

Calculate the mass of silver chloride required to plate 265 mg of pure silver.

Answer 1

0.35215 grams of silver chloride required to plate 265 mg of pure silver.

Step-by-step explanation:

`2AgCl(aq)\rightarrow 2Ag(s)+Cl_2(g)`

Mass of silver = 265 mg = 0.265 g

Moles of silver =
`(0.265 g)/(108 g/mol)=0.002454 mol`

According to reaction, 2 moles of silver are obtained from 2 moles of silver chloride.

Then 0.002454 moles of silver will be obtained from :

`(2)/(2)* 0.002454 mol=0.002454 mol` of silver chloride

Mass of 0.002454 moles of silver chloride:

= 0.002454 mol × 143.5 g/mol = 0.35215 g

0.35215 grams of silver chloride required to plate 265 mg of pure silver.

Answer 2

Given Mass of pure silver (Ag) = 265 mg

Silver chloride AgCl which is used in plating silver contains 75.27 % Ag

This means that:

A 100 mg of silver chloride contains 75.27 mg of silver

Therefore, the amount of silver chloride required to plate a sample containing 265 mg silver would correspond to:

265 mg Ag * 100 mg AgCl/75.27 mg Ag

= 352.1 mg AgCl