convert to gas vol vol = moles * 22.4 convert to particles particles = moles * 6.02 * 10 23 convert...
TRANSCRIPT
CONVERT TO GAS VOL VOL = MOLES * 22.4
CONVERT TO PARTICLES
PARTICLES = MOLES * 6.02 * 1023
CONVERT TO MOLES VOLUME = MOLES/ M
RECOGNIZE: MASSGIVEN 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 * 1023
RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVEUSE 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.
CALCULATE THE YEILD OF NH3 in grams IF 8 GRAMS OF H2 REACTS COMPLETLEY WITH EXCESS NITROGEN.
N2 + 3H2 2NH3
CONVERT TO GAS VOL VOL = MOLES * 22.4
CONVERT TO PARTICLES
PARTICLES = MOLES * 6.02 * 1023
CONVERT TO MOLES VOLUME = MOLES/ M
RECOGNIZE:
GIVEN IS A MASS IN GRAMS, ANY PHASE.
RECOGNIZE
GIVEN IS A GAS (g) VOLUME AT STP
RECOGNIZEGIVEN 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 * 1023
RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVEUSE 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.
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 H2 is a given mass.
CALCULATE THE YEILD OF NH3 in grams IF 8.0 GRAMS OF H2 REACTS COMPLETLEY WITH EXCESS NITROGEN.
N2 + 3H2 2NH3
MOLES = MASS/GFM MOLES = 8.00g/ 2.0 g/mol MOLES = 4.0 mol
STEP 2 RATIO known to objective, use reaction coefficients to compare H2 known) to NH3 (objective).
H2 = 3 = 4.0 mol , NH3 2 X X = 2.66 mol NH3
STEP 3 Convert Objective to required units. Convert ammonia from MOLES to GRAMS.
MOLES = MASS/GFM 2.66 = MASS/ 17.0 g/mol MASS = 45. g
CALCULATE THE REQUIRED VOLUME OF H2 GAS AT STP TO PRODUCE 40 GRAMS OF NH3?
N2 + 3H2 2NH3
CONVERT TO GAS VOL VOL = MOLES * 22.4
CONVERT TO PARTICLES
PARTICLES = MOLES * 6.02 * 1023
CONVERT TO MOLES VOLUME = MOLES/ M
RECOGNIZE:
GIVEN IS A MASS IN GRAMS, ANY PHASE.
RECOGNIZE
GIVEN IS A GAS (g) VOLUME AT STP
RECOGNIZEGIVEN 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 * 1023
RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVEUSE 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.
CALCULATE THE REQUIRED VOLUME OF H2 GAS AT STP TO PRODUCE 40.0 GRAMS OF NH3?
N2 + 3H2 2NH3
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 NH3 is a given mass.
MOLES = MASS/GFM MOLES = 40.0g/ 17.0 g/mol MOLES = 2.352 mol NH3
STEP 2 RATIO known to objective, use reaction coefficients to compare H2 known) to NH3 (objective).
H2 = 3 = X , NH3 2 2.352 X = 3.528 mol H2
STEP 3 Convert Objective to required units. Convert hydrogen from MOLES to STP gas Volume.. VOLUME = MOLES * 22.4L V = 3.528 mol H2 * 22.4L /mol V = 79.03 L
CALCULATE THE MASS OF HYDROGEN REQUIRED TO PRODUCE 67.2 L OF AMMONIA AT STP.
N2 + 3H2 2NH3
CONVERT TO GAS VOL VOL = MOLES * 22.4
CONVERT TO PARTICLES
PARTICLES = MOLES * 6.02 * 1023
CONVERT TO MOLES VOLUME = MOLES/ M
RECOGNIZE:
GIVEN IS A MASS IN GRAMS, ANY PHASE.
RECOGNIZE
GIVEN IS A GAS (g) VOLUME AT STP
RECOGNIZEGIVEN 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 * 1023
RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVEUSE 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.
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/GFM 4.5 MOLES = MASS/ 2.0 g/mol MASS = 9.00 GRAMS H2
STEP 2 RATIO known to objective, use reaction coefficients to compare H2 known) to NH3 (objective).
H2 = 3 = X , NH3 2 3 X = 4.50 mol H2
VOLUME = MOLES * 22.4L 67.2 = mol NH3 * 22.4L /mol MOL NH3 = 3
CALCULATE THE MASS OF HYDROGEN REQUIRED TO PRODUCE 67.2 L OF AMMONIA AT STP.
N2 + 3H2 2NH3
CALCULATE THE REQUIRED VOLUME OF N2 GAS AT STP TO PRODUCE 12.04 * 1023 MOLECULES OF
NH3?
N2(g) + 3H2(g) 2NH3(g)
CONVERT TO GAS VOL VOL = MOLES * 22.4
CONVERT TO PARTICLES
PARTICLES = MOLES * 6.02 * 1023
CONVERT TO MOLES VOLUME = MOLES/ M
RECOGNIZE:
GIVEN IS A MASS IN GRAMS, ANY PHASE.
RECOGNIZE
GIVEN IS A GAS (g) VOLUME AT STP
RECOGNIZEGIVEN 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 * 1023
RATIO MOLES GIVEN TO MOLES OBJECTIVE TO SOLVE FOR MOLES OF OBJECTIVEUSE 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.
CALCULATE THE REQUIRED VOLUME OF N2 GAS AT STP TO PRODUCE 12.04 * 1023 MOLECULES OF NH3?
N2(g) + 3H2(g) 2NH3(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 * 1023 MOLECULES (PARTICLES) OF NH3 IS THE KNOWN.
MOL= PART/6.02*1023 MOL=12.04*1023 / 6.02*1023 MOLES = 2.000 MOL NH3
STEP 2 RATIO known to objective, use reaction coefficients to compare N2 known) to NH3 (objective).
N2 = 1 = X , NH3 2 2.000 X = 1.000 mol N2
STEP 3 Convert Objective to required units. Convert hydrogen from MOLES to STP gas Volume.. VOLUME = MOLES * 22.4L V = 1.000 mol H2 * 22.4L /mol V = 22.40 L