Question #16281

Suppose that 32.9 mL of 0.479 M NaCl is added to 32.9 mL of 0.331 M AgNO3.
(a) How many moles of AgCl would precipitate?

Expert's answer

0.479 mol in 1000 mL

x & in& 32.9& mL

x=0.479*32.9/1000=0.01576 mol of NaCl

0.331 mol in 1000 ml

y & in& 32.9

y=0.0109 mol of AgNO3&

NaCl is in excess

NaCl+AgNO3---->NaNO3 + AgCl

It means that 0.0109 mol of AgNO3 can form 0.0109 mol& AgCl

x & in& 32.9& mL

x=0.479*32.9/1000=0.01576 mol of NaCl

0.331 mol in 1000 ml

y & in& 32.9

y=0.0109 mol of AgNO3&

NaCl is in excess

NaCl+AgNO3---->NaNO3 + AgCl

It means that 0.0109 mol of AgNO3 can form 0.0109 mol& AgCl

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