Question

I would like help understanding these questions.

4. Consider the following unbalanced equation: AgNO3(aq) + Cu(s) → Cu(NO3)2 (aq) + _ _Ag (s) (a) Balance the equation. (b) Calculate the amount of metallic silver (in grams) that can be produced from a solution containing 5.00 g of metallic copper and 20.0 g of silver(I) nitrate. (c) Calculate the amount of the reagent (in grams) of the excess reagent. (d) Calculate the percent yield if the reaction produces 11.5 g of metallic silver.

Thank you for your time!

Answer 1

a)

Balanced chemical equation is:

2 AgNO3(aq) + Cu(s) —> Cu(NO3)2(aq) + 2 Ag(s)

b)

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(AgNO3)= 20.0 g

use:

number of mol of AgNO3,

n = mass of AgNO3/molar mass of AgNO3

=(20 g)/(1.699*10^2 g/mol)

= 0.1177 mol

Molar mass of Cu = 63.55 g/mol

mass(Cu)= 5.0 g

use:

number of mol of Cu,

n = mass of Cu/molar mass of Cu

=(5 g)/(63.55 g/mol)

= 7.868*10^-2 mol

2 mol of AgNO3 reacts with 1 mol of Cu

for 0.1177 mol of AgNO3, 5.885*10^-2 mol of Cu is required

But we have 7.868*10^-2 mol of Cu

so, AgNO3 is limiting reagent

we will use AgNO3 in further calculation

Molar mass of Ag = 107.9 g/mol

According to balanced equation

mol of Ag formed = (2/2)* moles of AgNO3

= (2/2)*0.1177

= 0.1177 mol

use:

mass of Ag = number of mol * molar mass

= 0.1177*1.079*10^2

= 12.7 g

Answer: 12.7 g

c)

According to balanced equation

mol of Cu reacted = (1/2)* moles of AgNO3

= (1/2)*0.1177

= 5.885*10^-2 mol

mol of Cu remaining = mol initially present - mol reacted

mol of Cu remaining = 7.868*10^-2 - 5.885*10^-2

mol of Cu remaining = 1.982*10^-2 mol

Molar mass of Cu = 63.55 g/mol

use:

mass of Cu,

m = number of mol * molar mass

= 1.982*10^-2 mol * 63.55 g/mol

= 1.26 g

Answer: 1.26 g

d)

% yield = actual mass*100/theoretical mass

= 11.5*100/12.7

= 90.6 %

Answer: 90.6 %