Question

How many grams of silver are produced when 3.00g of silver nitrate react with 10g of copper?

Answer 1

Molar mass of AgNO3 = 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 = 3.0 g

we have below equation to be used:

number of mol of AgNO3,

n = mass of AgNO3/molar mass of AgNO3

=(3.0 g)/(169.91 g/mol)

= 1.766*10^-2 mol

Molar mass of Cu = 63.55 g/mol

mass of Cu = 10.0 g

we have below equation to be used:

number of mol of Cu,

n = mass of Cu/molar mass of Cu

=(10.0 g)/(63.55 g/mol)

= 0.1574 mol

we have the Balanced chemical equation as:

2 AgNO3 + Cu ---> 2 Ag + Cu(NO3)2

2 mol of AgNO3 reacts with 1 mol of Cu

for 1.766*10^-2 mol of AgNO3, 8.828*10^-3 mol of Cu is required

But we have 0.1574 mol of Cu

so, AgNO3 is limiting reagent

we will use AgNO3 in further calculation

Molar mass of Ag = 107.9 g/mol

From balanced chemical reaction, we see that

when 2 mol of AgNO3 reacts, 2 mol of Ag is formed

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

= (2/2)*1.766*10^-2

= 1.766*10^-2 mol

we have below equation to be used:

mass of Ag = number of mol * molar mass

= 1.766*10^-2*1.079*10^2

= 1.905 g

Answer: 1.90 g