NEED THIS FOR A QUIZ

Question

For the following reaction, 162 grams of silver nitrate are allowed to react with 25.2 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

To solve it, first we need to write down a balanced chemical equation. Since nitrate has a charge of -1, and silver +1, silver nitrate is AgNO₃, while Copper (II) Nitrate is CuNO₃.

So the balanced equation is 2AgNO₃ + Cu -> Cu(NO₃)₂ + 2Ag

The main takeaway is that for every mole of Copper that goes in, 2 moles of silver nitrate goes in.

Now lets find how many moles there are of Copper and Silver Nitrate.

The molar mass of copper is 63.5 (to 3 sig figs), so there are 25.2/63.5 = 0.397 mols of copper

The molar mass of silver nitrate is 170. so there are 162/170 = 0.953 mols of silver nitrate.

Since for every mole of copper that goes in, two moles of silver nitrate goes in, to use up all the copper we need 0.397*2 = 0.794 mols of silver nitrate. Since we have more than that (0.953>0.794), the copper runs out first and Copper is the limiting reagent.

Looking at the equation, for every mole of copper that goes in, 1 mole of copper (II) nitrate comes out. So the number of moles of copper (II) nitrate is equal to the number of moles of copper which is 0.397 moles.

To find the mass, multiply by the molar mass of copper (II) nitrate which is 188. so you find the mass of Copper (II) Nitrate to be 74.6 grams.

Since we only used up 0.794 moles of silver nitrate, there are 0.953-0.794 = 0.159 moles remaining. Multiply this by the molar mass which is 170 and you find that the mass of the excess reagent is 27.0 grams.