The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4g of `AgNO_(3)`, is
[Atomie mass of Ag -108, Atomic mass of Na - 23]

Options

A. 5.74 g
B. 1.17 g
C. 2.87 g
D. 6.8 g

Answer

C

Explanation

(c) : `NaCl + AgNO_(3) to AgCl+NaNO_(3)`
No. of moles : `(11.70)/(58.5) (3.4)/(170)`
= 0.2 = 0.02 = 0.02
Here, `AgNO_(3)` is limiting reagent.
`:.` Mass of AgCl precipitated = `0.02 xx 143.5 = 2.87` g