Now that we understand the role of coefficients in chemical equations, we can convert between grams and moles to determine the mass of the products that result from a chemical reaction. Let’s conduct an experiment where we combine copper and silver nitrate.

Here is the following balanced chemical reaction:

Cu(s) + 2AgNO_{3}(*aq*) → Cu(NO_{3})_{2}(*s*) + 2Ag(s)

First off, let’s notice this is a single replacement reaction.

AB + C → A + CB, where Cu = A, NO_{3} = B, and Ag = C

**Question:**Find the mole ratio of AgNO_{3}to Cu(NO_{3})_{2}**Answer:**2:1**Question:**Find the mole ration of AgNO_{3}to Ag**Answer:**1:1- Suppose you have 6.0 mol of Cu.
**Question:**How many moles of AgNO_{3 }are needed to react completely with 6.0 mol of Cu?**Answer:**The mole ratio is 2 moles of AgNO_{3 }to 1 mole of Cu, so to completely react 6.0 mol of Cu, you need twice as many moles of AgNO_{3}(12 mol).**Question:**How many moles of Cu(NO_{3})_{2}are produced?**Answer:**The mole ratio of Cu(NO_{3})_{2 }to Cu is 1:1, so if you have 6.0 mol of Cu, you produce 6.0 mol of Cu(NO_{3})_{2}.

- Suppose you want to make 30.0g of Ag.
**Question:**How many moles of Ag is that?**Answer:**The molar mass of Ag is 107.9 g/mol. 30g x (1mol / 107.9g) = 0.278 mol Ag**Question:**How many moles of Cu do you need?**Answer:**The mole ration of Ag to Cu is 2:1, so with 0.278 mol Ag, you would need 0.139 mol of Cu.

Imagine we actually want to do this single replacement experiment where we produce 30.0 g of silver. You go to the chemical supply shelf and locate the bottle of copper and silver nitrate. Both are in solid form. Now what? The Mole Tunnel! In the real world, we work in mass quantities like grams. To move through a chemical reaction from products to reactants (or vice versa), you go from grams to moles to grams.

We know we need 0.139 mol of Cu. The copper:silver nitrate ratio is 1:2, so we need twice as many moles of silver nitrate, or 0.278 mol AgNO_{3}. To actually do this experiment, we need to use counting by weighing, which means converting moles to grams. To convert moles to grams, we need molar mass (g/mol). For Cu, the molar mass is 63.55 g/mol. For AgNO_{3}, the molar mass is 169.9 g/mol.

- 0.139 mol Cu x 63.55 g/mol = 8.83 g of Cu
- 0.278 mol AgNO
_{3}x 169.9 g/mol = 47.2 g of AgNO_{3}

Done! You successfully navigated the Mole Tunnel! You started with grams (30 g), converted to moles, compared mole ratios, then converted back to grams. By combining 8.83 g of Cu with 47.2 g of AgNO_{3}, you will produce 30 g of Ag. Math!

Your turn! Complete the Mole Tunnel Google Form before returning to Week 33 – Stoichiometry to continue working.