Question

1 gram of silver ore is dissolved in hno3 and ag precipitated by hcl. the mass obtained agcl is 0.7880g calculate the % amount of ag in the ore. Also don’t spam for points, I will delete ur answer and report u

Answer 1

Explanation:

Answer: 49.9 % of silver was present in a silver coin.

Explanation:

AgNO_3+NaCl\rightarrow AgCl+NaNO_3

Moles of AgCl = \frac{\text{mass of AgCl}}{\text{MOlar mass of AgCl}}=\frac{14.35 g}{143.32 g/mol}=0.1001 mol

in one molecular formula of AgCl there are one Ag atom then in 0.1001 mol of AgCl will contain 0.1001 mole of Ag metal.

Mass of Ag metal = Moles of Ag × Molar mass of Ag

=0.1001 moles\times 107.87 g/mol=10.79 g

\%\text{of Ag}=\frac{\text{mass of Ag}}{\text{total mass of silver coin}}\times 100=\frac{10.79 g}{21.6 g}\times 100=49.9\%

The weight of AgCl is 14.35g then percentage of silver in coin is 49.9 %.