Question

What is mass of precipitate formed when 50 ml of 16.9%solution of AgNO3 is mixed with 50ml of 5.8% of NaCl solution

Answer 1

heya...


moles of AgNo3= 50x 16.9/100x169.8

                           = 0.05mole

Moles of NaCl = 50x5.8/100x58.5

                       =0.05mole

AgNo3 + NaCl ----> AgCl + NaNo3

mass of AgCl = mole x molar mass

                       =0.05 x 143.5

                       = 7.16 g

tysm.#gozmit

Answer 2

Hey mate!

Here's your answer!!

Moles of AgNO3
= > 50 × 16.9/(100 * 170)
=> 0.05 (approx.)

Moles of NaCl
= > 5.8 * 5.8/(100 * 58.5)
=> 0.05

The reaction goes like this:
AgNO3 + NaCl ---------» AgCl + NaNO3
0.05 0.05 0
0
0 0 0.05

Mass of AgCl
= 0.05 143.5
=> 7.16 g.

•Hence this would be the final mass of the precipitate.

$hope \: it \: helps \: you.$
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