1. Another Practice Problem

2. You walk into a room and find two chemical bottles open and sitting out. One contains lead (II) nitrate and the other potassium chloride. You also find a beaker containing 50 grams of lead (II) chloride Pb(NO 3 ) 2 + KCl  PbCl 2 + KNO 3

3. How much lead (II) nitrate and potassium chloride did the person combine to make 50 grams of lead (II) chloride? Step 1- Balance it Pb(NO 3 ) 2 + 2KCl  PbCl 2 + 2KNO 3

4. Step 2- Determine the molar mass of PbCl 2 Pb = 207.2 g/mol 2Cl = 2(35.45 g/mol) PbCl 2 = 278.1 g/mol

5. Pb(NO 3 ) 2 + 2KCl  PbCl 2 + 2KNO 3 Step 3- Convert grams to moles 50 g = 0.18 mol of PbCl 2 278.1 g/mol

6. Pb(NO 3 ) 2 + 2KCl  PbCl 2 + 2KNO 3 Step 4- Use the mole ratios to figure out how many moles of each substance were used and created..18 mol PbCl 2 1 Pb(NO 3 ) 2 : 2 KCl : 1 PbCl 2 : 2 KNO 3.18Pb(NO 3 ) 2 +.36KCl .18PbCl 2 +.36KNO 3

7. Pb(NO 3 ) 2 + 2KCl  PbCl 2 + 2KNO 3 Step 5 – Determine the molar mass of both reactants Pb(NO 3 ) 2 = 331.2 g/mol KCl = 74.55 g/mol

8. Pb(NO 3 ) 2 + 2KCl  PbCl 2 + 2KNO 3 Step 6 – convert moles to grams for each reactant Pb(NO 3 ) 2 = 331.2 g/mol *.18 mol = 51.62 g KCl = 74.55 g/mol *.36 mol = 26.84 g