1. CONVERT TO GAS VOL VOL = MOLES * 22.4 CONVERT TO PARTICLES PARTICLES = MOLES * 6.02 * 10 23 CONVERT TO MOLES VOLUME = MOLES/ M RECOGNIZE: MASS GIVEN IS A MASS IN GRAMS, ANY PHASE. RECOGNIZE: STP GAS GIVEN IS A GAS (g) VOLUME AT STP RECOGNIZE GIVEN IS IN PARTICLES (ATOMS OR MOLECULES) RECOGNIZE: VOLUME AND MOLARITY GIVEN IS AN AQUIOUS SOLUTION (aq) CONVERT TO MOLES MOLES = MASS/GFM CONVERT TO MOLES MOLES =VOL(L)/22.4L CONVERT TO MOLES MOLES = PARTICLES 6.02 * 10 23 RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVE USE COEFFICIENT OF BALANCED REACTION CONVERT TO MOLES MOLES =VOLUME * M CONVERT TO MASS MASS = MOLES * GFM STEP ONE: CONVERT KNOWN TO MOLES. STEP TWO: MOLE RATIO STEP THREE: CONVERT OBJECTIVE TO UNITS.

2. CALCULATE THE YEILD OF NH 3 in grams IF 8 GRAMS OF H 2 REACTS COMPLETLEY WITH EXCESS NITROGEN. N 2 + 3H 2  2NH 3 CONVERT TO GAS VOL VOL = MOLES * 22.4 CONVERT TO PARTICLES PARTICLES = MOLES * 6.02 * 10 23 CONVERT TO MOLES VOLUME = MOLES/ M RECOGNIZE: GIVEN IS A MASS IN GRAMS, ANY PHASE. RECOGNIZE GIVEN IS A GAS (g) VOLUME AT STP RECOGNIZE GIVEN IS IN PARTICLES (ATOMS OR MOLECULES) RECOGNIZE GIVEN IS AN AQUIOUS SOLUTION (aq) CONVERT TO MOLES MOLES = MASS/GFM CONVERT TO MOLES MOLES =VOL(L)/22.4L CONVERT TO MOLES MOLES = PARTICLES 6.02 * 10 23 RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVE USE COEFFICIENT OF BALANCED REACTION CONVERT TO MOLES MOLES =VOLUME * M CONVERT TO MASS MASS = MOLES * GFM STEP ONE: CONVERT KNOWN TO MOLES. STEP TWO: MOLE RATIO STEP THREE: CONVERT OBJECTIVE TO UNITS.

3. STEP 1 Identify known (the substance that can be converted to moles), choose the correct equation to convert to moles. In this problem 8.0 grams of H 2 is a given mass. CALCULATE THE YEILD OF NH 3 in grams IF 8.0 GRAMS OF H 2 REACTS COMPLETLEY WITH EXCESS NITROGEN. N 2 + 3H 2  2NH 3 MOLES = MASS/GFMMOLES = 8.00g/ 2.0 g/molMOLES = 4.0 mol STEP 2 RATIO known to objective, use reaction coefficients to compare H 2 known) to NH 3 (objective). H 2 = 3 = 4.0 mol, NH 3 2 X X = 2.66 mol NH 3 STEP 3 Convert Objective to required units. Convert ammonia from MOLES to GRAMS. MOLES = MASS/GFM2.66 = MASS/ 17.0 g/molMASS = 45. g

4. CALCULATE THE REQUIRED VOLUME OF H 2 GAS AT STP TO PRODUCE 40 GRAMS OF NH 3 ? N 2 + 3H 2  2NH 3 CONVERT TO GAS VOL VOL = MOLES * 22.4 CONVERT TO PARTICLES PARTICLES = MOLES * 6.02 * 10 23 CONVERT TO MOLES VOLUME = MOLES/ M RECOGNIZE: GIVEN IS A MASS IN GRAMS, ANY PHASE. RECOGNIZE GIVEN IS A GAS (g) VOLUME AT STP RECOGNIZE GIVEN IS IN PARTICLES (ATOMS OR MOLECULES) RECOGNIZE GIVEN IS AN AQUIOUS SOLUTION (aq) CONVERT TO MOLES MOLES = MASS/GFM CONVERT TO MOLES MOLES =VOL(L)/22.4L CONVERT TO MOLES MOLES = PARTICLES 6.02 * 10 23 RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVE USE COEFFICIENT OF BALANCED REACTION CONVERT TO MOLES MOLES =VOLUME * M CONVERT TO MASS MASS = MOLES * GFM STEP ONE: CONVERT KNOWN TO MOLES. STEP TWO: MOLE RATIO STEP THREE: CONVERT OBJECTIVE TO UNITS.

5. CALCULATE THE REQUIRED VOLUME OF H 2 GAS AT STP TO PRODUCE 40.0 GRAMS OF NH 3 ? N 2 + 3H 2  2NH 3 STEP 1 Identify known (the substance that can be converted to moles), choose the correct equation to convert to moles. In this problem 40 grams of NH 3 is a given mass. MOLES = MASS/GFMMOLES = 40.0g/ 17.0 g/molMOLES = 2.352 mol NH 3 STEP 2 RATIO known to objective, use reaction coefficients to compare H 2 known) to NH 3 (objective). H 2 = 3 = X, NH 3 2 2.352 X = 3.528 mol H 2 STEP 3 Convert Objective to required units. Convert hydrogen from MOLES to STP gas Volume.. VOLUME = MOLES * 22.4LV = 3.528 mol H 2 * 22.4L /molV = 79.03 L

6. CALCULATE THE MASS OF HYDROGEN REQUIRED TO PRODUCE 67.2 L OF AMMONIA AT STP. N 2 + 3H 2  2NH 3 CONVERT TO GAS VOL VOL = MOLES * 22.4 CONVERT TO PARTICLES PARTICLES = MOLES * 6.02 * 10 23 CONVERT TO MOLES VOLUME = MOLES/ M RECOGNIZE: GIVEN IS A MASS IN GRAMS, ANY PHASE. RECOGNIZE GIVEN IS A GAS (g) VOLUME AT STP RECOGNIZE GIVEN IS IN PARTICLES (ATOMS OR MOLECULES) RECOGNIZE GIVEN IS AN AQUIOUS SOLUTION (aq) CONVERT TO MOLES MOLES = MASS/GFM CONVERT TO MOLES MOLES =VOL(L)/22.4L CONVERT TO MOLES MOLES = PARTICLES 6.02 * 10 23 RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVE USE COEFFICIENT OF BALANCED REACTION CONVERT TO MOLES MOLES =VOLUME * M CONVERT TO MASS MASS = MOLES * GFM STEP ONE: CONVERT KNOWN TO MOLES. STEP TWO: MOLE RATIO STEP THREE: CONVERT OBJECTIVE TO UNITS.

7. STEP 3 Convert Objective to required units. Convert hydrogen from MOLES to MASS. STEP 1 Identify known (the substance that can be converted to moles), choose the correct equation to convert to moles. In this problem 67.2 L OF AMMONIA IS GIVEN(KNOWN) GAS VOLUME. MOLES = MASS/GFM4.5 MOLES = MASS/ 2.0 g/molMASS = 9.00 GRAMS H 2 STEP 2 RATIO known to objective, use reaction coefficients to compare H 2 known) to NH 3 (objective). H 2 = 3 = X, NH 3 2 3 X = 4.50 mol H 2 VOLUME = MOLES * 22.4L67.2 = mol NH 3 * 22.4L /molMOL NH 3 = 3 CALCULATE THE MASS OF HYDROGEN REQUIRED TO PRODUCE 67.2 L OF AMMONIA AT STP. N 2 + 3H 2  2NH 3

8. CALCULATE THE REQUIRED VOLUME OF N 2 GAS AT STP TO PRODUCE 12.04 * 10 23 MOLECULES OF NH 3 ? N 2(g) + 3H 2(g)  2NH 3(g) CONVERT TO GAS VOL VOL = MOLES * 22.4 CONVERT TO PARTICLES PARTICLES = MOLES * 6.02 * 10 23 CONVERT TO MOLES VOLUME = MOLES/ M RECOGNIZE: GIVEN IS A MASS IN GRAMS, ANY PHASE. RECOGNIZE GIVEN IS A GAS (g) VOLUME AT STP RECOGNIZE GIVEN IS IN PARTICLES (ATOMS OR MOLECULES) RECOGNIZE GIVEN IS AN AQUIOUS SOLUTION (aq) CONVERT TO MOLES MOLES = MASS/GFM CONVERT TO MOLES MOLES =VOL(L)/22.4L CONVERT TO MOLES MOLES = PARTICLES 6.02 * 10 23 RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVE USE COEFFICIENT OF BALANCED REACTION CONVERT TO MOLES MOLES =VOLUME * M CONVERT TO MASS MASS = MOLES * GFM STEP ONE: CONVERT KNOWN TO MOLES. STEP TWO: MOLE RATIO STEP THREE: CONVERT OBJECTIVE TO UNITS.

9. CALCULATE THE REQUIRED VOLUME OF N 2 GAS AT STP TO PRODUCE 12.04 * 10 23 MOLECULES OF NH 3 ? N 2(g) + 3H 2(g)  2NH 3(g) STEP 1 Identify known (the substance that can be converted to moles), choose the correct equation to convert to moles. In this problem 12.04 * 10 23 MOLECULES (PARTICLES) OF NH 3 IS THE KNOWN. MOL= PART/6.02*10 23 MOL=12.04*10 23 / 6.02*10 23 MOLES = 2.000 MOL NH 3 STEP 2 RATIO known to objective, use reaction coefficients to compare N 2 known) to NH 3 (objective). N 2 = 1 = X, NH 3 2 2.000 X = 1.000 mol N 2 STEP 3 Convert Objective to required units. Convert hydrogen from MOLES to STP gas Volume.. VOLUME = MOLES * 22.4LV = 1.000 mol H 2 * 22.4L /molV = 22.40 L