Question

Zinc metal and aqueous silver nitrate react according to the equation:

Zn(s) + 2AgNO3 ---> 2Ag (s) + Zn(NO3)2

When an excess of Zn reacts with 25.00 g of AgNO₃, (1.30x10^1) grams of silver are produced. What is the percent yield?

Answer 1

Molar mass of AgNO3,

MM = 1*MM(Ag) + 1*MM(N) + 3*MM(O)

= 1*107.9 + 1*14.01 + 3*16.0

= 169.91 g/mol

mass of AgNO3 = 25 g

mol of AgNO3 = (mass)/(molar mass)

= 25/1.699*10^2

= 0.1471 mol

Balanced chemical equation is:

Zn(s) + 2AgNO3 ---> 2Ag (s) + Zn(NO3)2

According to balanced equation

mol of Ag formed = moles of AgNO3

= 0.1471 mol

Molar mass of Ag = 107.9 g/mol

mass of Ag = number of mol * molar mass

= 0.1471*1.079*10^2

= 15.88 g

% yield = actual mass*100/theoretical mass

= 13.0*100/15.88

= 81.9 %

Answer: 81.9 %