Equal volumes of 0.140 M AgNO3 and 0.200 M ZnCl2 solution are mixed. Calculate the equilibrium concentrations of Ag+ and Zn2+.

The concentration of Zn²⁺ = 0.065M

Explanation:

2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂

According to the balanced equation, 2 moles of AgNO₃ reacts with 1 mole of ZnCl₂ to form 2 moles of AgCl and 1 mole of Zn(NO₃)₂.

Molarity of AgNO₃ is 0.14M

Molarity of ZnCl₂ is 0.2M

0.14 M of AgNO₃ would react completely with 0.14/2) = 0.07 M of ZnCl₂, but the ZnCl₂ is more concentrated than that, so ZnCl₂ is in excess and AgNO₃ is the limiting reactant.

Neglecting the Ksp of AgCl:

All of the Ag is in the precipitate so the concentration of Ag⁺ is zero.

(0.2 M ZnCl2 originally) - (0.07 M ZnCl2 reacted) = 0.13 M ZnCl₂ = 0.13 M Zn²⁺ of the original ZnCl₂ solution.

But the original solution has been diluted by the addition of an equal  volume of AgNO₃, so the concentration in the final solution is one-half of the original, so:

0.13 M Zn²⁺ / 2

= 0.065 M Zn²⁺

AgCl(s) is also produced as a precipitate so the ions comprising AgCl are removed from solution.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

what is the question?