CHEMISTRY HOMEWORK HELP

Question

For the following reaction, 115 grams of silver nitrate are allowed to react with 25.4 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?

What is the FORMULA for the limiting reagent?

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

Answer 1

One of the easiest ways to identify the limiting reagent is to divide moles of each reactant by its coefficient in the balanced equation, and which ever is lower is the limiting reactant. In the current situation we have...

2AgNO₃(aq) + Cu(s) ==> Cu(NO₃)₂(aq) + 2Ag(s) ... balanced equation

moles AgNO₃ = 115 g x 1mol/170 g = 0.6765 moles

moles Cu = 25.4 g x 1 mol/63.55 g = 0.3997 moles

If we divide 0.6765 by 2 we get 0.338 for AgNO₃. Dividing 0.3997 by 1 = 0.3997. So, AgNO₃ is limiting.

Maximum amount Cu(NO₃)₂:

0.677 moles AgNO₃ x 1 mol Cu(NO₃)₂ / 2 mol AgNO₃ = 0.339 moles Cu(NO₃)₂. If you want grams, multiply by the molar mass

Excess remaining:

0.677 mol AgNO₃ x 1 mol Cu/2 mol AgNO₃ = 0.339 moles Cu used up

0.3997 moles initially present - 0.339 moles used = 0.0607 moles remaining

0.0607 moles Cu x 63.55 g/mol = 3.86 g Cu remaining