Answer to Question #344971 in General Chemistry for gle

Question

For the following reaction, 108 grams of silver nitrate are allowed to react with 23.6 grams of copper .

silver nitrate(aq) + copper(s)  copper(II) nitrate(aq) + silver(s)

What is the maximum amount of copper(II) nitrate that can be formed?  grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?  grams

Answer 1

Solution:

The molar mass of silver nitrate (AgNO₃) is 169.87 g/mol

The molar mass of copper (Cu) is 63.546 g/mol

Calculate moles of each reagent:

Moles of AgNO₃ = (108 g AgNO₃) × (1 mol AgNO₃ / 169.87 g AgNO₃) = 0.6358 mol AgNO₃

Moles of Cu = (23.6 g Cu) × (1 mol Cu / 63.546 g Cu) = 0.3714 mol Cu

Balanced chemical equation:

2AgNO₃(aq) + Cu(s) → Cu(NO₃)₂(aq) + 2Ag(s)

According to stoichiometry:

2 mol of AgNO₃ react with 1 mol of Cu

Thus, 0.6358 mol of AgNO₃ react with:

(0.6358 mol AgNO₃) × (1 mol Cu / 2 mol AgNO₃) = 0.3179 mol Cu

However, initially there is 0.3714 mol of Cu (according to the task)

Therefore, AgNO₃ acts as limiting reagent and Cu is excess reagent

According to stoichiometry:

2 mol of AgNO₃ produce 1 mol of Cu(NO₃)₂

Thus, 0.6358 mol of AgNO₃ produce:

(0.6358 mol AgNO₃) × (1 mol Cu(NO₃)₂ / 2 mol AgNO₃) = 0.3179 mol Cu(NO₃)₂

The molar mass of copper(II) nitrate (Cu(NO₃)₂) is 187.56 g/mol

Therefore,

Mass of Cu(NO₃)₂ = [ 0.3179 mol Cu(NO₃)₂] × [187.56 g Cu(NO₃)₂ / 1 mol Cu(NO₃)₂] = 59.6 g Cu(NO₃)₂

Mass of Cu(NO₃)₂ is 59.6 grams

Moles of Cu after the reaction = (0.3714 mol − 0.3179 mol) = 0.0535 mol

Therefore,

Mass of Cu after the reaction = (0.0535 mol Cu) × (63.546 g Cu / 1 mol Cu) = 3.4 g Cu

Mass of Cu after the reaction is 3.4 grams

Answers:

The maximum amount of copper(II) nitrate is 59.6 grams

The limiting reagent is silver nitrate (AgNO₃)

3.4 grams of copper (Cu) remain after the reaction