The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4g of `AgNO_3` is

Options

A. 28g
B. 9.25g
C. 2.870g
D. 58g

Answer

C

Explanation

`AgNO_3+NaCl to AgCl + NaNO_3`
No. of moles of `AgNO_3=(3.4)/(170)=0.02`
No. of moles of NaCl `=(5.85)/(58.5)=0.1`
limiting reagent = `AgNO_3`
1 mole of `AgNO_3` produces 1 mole of AgCl
0.02 mole of `AgCO_3` produce 0.02 mole of AgCl
Weight of AgCl produced = `0.02xx143.5=2.870 g`