quantities in chemical reactions review definitions $100 $200 $300 $400 $500 quantities balanced...
TRANSCRIPT
Quantities in Chemical Reactions
Review
Quantities in Chemical Reactions
Review
Definitions
$100 $100 $100 $100 $100 $100
$200 $200 $200 $200 $200
$300 $300 $300 $300 $300 $300
$400 $400 $400 $400 $400 $400
$500 $500 $500 $500 $500 $500
QuantitiesBalanced Chemical Equations
Additional Calculations
Team 1 Team 2 Team 3 Team 4
$200
Moles & Mass
Empirical Formulas
Definitions: $100Definitions: $100
Answer
A unit of measurement expressing amount of
a species
Definitions: $100Definitions: $100
MoleMole
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Definitions: $200Definitions: $200
Answer
The relationship between amounts of species in a reaction
Definitions: $200Definitions: $200
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StoichiometryStoichiometry
Definitions: $300Definitions: $300
Answer
The mass of one mole of a particular
chemical species
Definitions: $300Definitions: $300
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Atomic MassAtomic Mass
Definitions: $400Definitions: $400
Answer
1 mole of any gas occupies 22.4 L
Definitions: $400Definitions: $400
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STPSTP
Definitions: $500Definitions: $500
Answer
The number of particles in one mole
of a species
Definitions: $500Definitions: $500
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Avogadro’s NumberAvogadro’s Number
Moles & Mass: $100Moles & Mass: $100
Answer
Molar mass of (NH 4)2SO4
Moles & Mass: $100Moles & Mass: $100
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132.14 g/mol
Moles & Mass: $200Moles & Mass: $200
Answer
Number of atoms in 0.00511 mol of NaNO3
Moles & Mass: $200Moles & Mass: $200
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1.539 * 1023 atoms
Moles & Mass: $300Moles & Mass: $300
Answer
Number of grams in 2.3 x 10-4 moles of calcium
phosphate, Ca3(PO3)2?
Moles & Mass: $300Moles & Mass: $300
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0.116 grams
Scores
Proceed
Moles & Mass: $400Moles & Mass: $400
Answer
Using your knowledge of mole calculations and unit conversions,
determine how many atoms there are in 1 gallon of gasoline. Assume that the molecular formula for gasoline is
C6H14 and that the density of gasoline is approximately 0.85 grams/mL.
Note: there are 3785 mL per gallon.
Moles & Mass: $400Moles & Mass: $400
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5.29067 * 1023 atoms
Moles & Mass: $500Moles & Mass: $500
Answer
Number of molecules are there in 450 grams of
Na2SO4
Moles & Mass: $500Moles & Mass: $500
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1.91 * 1024 molecules
Quantities: $100Quantities: $100
Answer
CH4 + 3 O2 CO2 + 2 H2
Moles of CO2 produced if 26 g of CH4 is burned
Quantities: $100Quantities: $100
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16 moles16 moles
Quantities: $200Quantities: $200
Answer
CH4 + 3 O2 CO2 + 2 H2
Moles of CO2 produced if 67.2 L of CH4 is burned
Quantities: $200Quantities: $200
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3 moles3 moles
Quantities: $300Quantities: $300
Answer
CaCO3(s) CO2(g) + CaO(s)
How many grams of calcium carbonate would you need to make 3.45 L of carbon dioxide
(assume STP)?
Quantities: $300Quantities: $300
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100/6 moles100/6 moles
Quantities: $400Quantities: $400
Answer
H2 gas can be produced by reacting CH4 with high
temperature steam. How many particles of H2 are produced when 158 g of CH4 reacts with steam?
CH4(g) + H2O(g) → CO(g) + 3 H2(g)
Quantities: $400Quantities: $400
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180 * 1023 = 1.8 * 1025180 * 1023 = 1.8 * 1025
Quantities: $500Quantities: $500
Answer
The Haber process combines hydrogen (H2) with nitrogen (N2) to produce ammonia (NH3). At
STP, how many moles of ammonia can you produce, given
that you have 44.8 L of H2?
N2 (g) +3H2(g) → 2NH3(g)
Quantities: $500Quantities: $500
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1.333 moles1.333 moles
Formulas: $100Formulas: $100
Answer
The empirical formulas, given the % composition for respective elements:
Fe = 63.53%, S = 36.47%
The empirical formulas, given the % composition for respective elements:
Fe = 63.53%, S = 36.47%
Formulas: $100Formulas: $100
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FeSFeS
Formulas: $200Formulas: $200
Answer
The empirical formulas, given the % composition for respective elements:
Na = 21.6%, Cl = 33.3%, O = 45.1%
The empirical formulas, given the % composition for respective elements:
Na = 21.6%, Cl = 33.3%, O = 45.1%
Formulas: $200Formulas: $200
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NaClO3NaClO3
Formulas: $300Formulas: $300
Answer
The empirical formulas, given the % composition for respective elements:
Cr = 26.52%, S = 24.52%, O = 48.96%
The empirical formulas, given the % composition for respective elements:
Cr = 26.52%, S = 24.52%, O = 48.96%
Formulas: $300Formulas: $300
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CrS3O6CrS3O6
Formulas: $400Formulas: $400
Answer
A compound of unknown identity is
made up of 40.1% sulfur and 59.9% oxygen (by
mass).
Formulas: $400Formulas: $400
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SO3SO3
Formulas : $500Formulas : $500
Answer
The empirical formulas, given the % composition for respective elements:
C = 63.1%, H = 11.92%, F = 24.97%
The empirical formulas, given the % composition for respective elements:
C = 63.1%, H = 11.92%, F = 24.97%
Formulas: $500Formulas: $500
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C5H12F
Balanced Chemical Equations: $100Balanced Chemical Equations: $100
Answer
__CHCl3 + __Cl2 → __CCl4 + __HCl __CHCl3 + __Cl2 → __CCl4 + __HCl
Balanced Chemical Equations: $100Balanced Chemical Equations: $100
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Already balanced
CHCl3 + Cl2 → CCl4 + HCl
Already balanced
CHCl3 + Cl2 → CCl4 + HCl
Balanced Chemical Equations: $200Balanced Chemical Equations: $200
Answer
__SnO2 + __H2 → __Sn + __H2O__SnO2 + __H2 → __Sn + __H2O
Balanced Chemical Equations: $200Balanced Chemical Equations: $200
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SnO2 + 2H2 → Sn + 2H2OSnO2 + 2H2 → Sn + 2H2O
Balanced Chemical Equations: $300Balanced Chemical Equations: $300
Answer
__KOH + __H3PO4→__K3PO4 + __H2O__KOH + __H3PO4→__K3PO4 + __H2O
Balanced Chemical Equations: $300Balanced Chemical Equations: $300
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3 KOH + H3PO4→ K3PO4 + 3 H2O3 KOH + H3PO4→ K3PO4 + 3 H2O
Balanced Chemical Equations: $400Balanced Chemical Equations: $400
Answer
__KNO3 + __H2CO3→__K2CO3 + __HNO3__KNO3 + __H2CO3→__K2CO3 + __HNO3
Balanced Chemical Equations: $400Balanced Chemical Equations: $400
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2 KNO3 + H2CO3 → K2CO3 + 2 HNO32 KNO3 + H2CO3 → K2CO3 + 2 HNO3
Balanced Chemical Equations: $500Balanced Chemical Equations: $500
Answer
__ CaCl2 + __ Na3PO4 → __ Ca3(PO4)2 + __ NaCl
__ CaCl2 + __ Na3PO4 → __ Ca3(PO4)2 + __ NaCl
Balanced Chemical Equations: $500Balanced Chemical Equations: $500
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3 CaCl2 + 2 Na3PO4 → Ca3(PO4)2 + 6 NaCl
3 CaCl2 + 2 Na3PO4 → Ca3(PO4)2 + 6 NaCl
Additional Calculations: $100Additional Calculations: $100
Answer
How many grams of H2 are produced if there are 18
grams of H2O?
CH4(g) + H2O(g) → CO(g) + 3 H2(g)
How many grams of H2 are produced if there are 18
grams of H2O?
CH4(g) + H2O(g) → CO(g) + 3 H2(g)
Additional Calculations: $100Additional Calculations: $100
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3 grams3 grams
Additional Calculations: $200Additional Calculations: $200
Answer
Na2S(aq) + AgNO3(aq) → Ag2S(s) + NaNO3(aq)
How many grams of Ag2S can be produced from 7.88 grams of AgNO3 and excess Na2S?
Na2S(aq) + AgNO3(aq) → Ag2S(s) + NaNO3(aq)
How many grams of Ag2S can be produced from 7.88 grams of AgNO3 and excess Na2S?
Additional Calculations: $200Additional Calculations: $200
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5.74 grams5.74 grams
Additional Calculations: $300Additional Calculations: $300
Answer
CuSO4 + Zn → ZnSO4 + Cu
How many grams of copper are produced from 2.9 grams
of zinc consumed with excess CuSO4 in this
reaction?
CuSO4 + Zn → ZnSO4 + Cu
How many grams of copper are produced from 2.9 grams
of zinc consumed with excess CuSO4 in this
reaction?
Additional Calculations: $300Additional Calculations: $300
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2.8 grams2.8 grams
Additional Calculations: $400Additional Calculations: $400
Answer
Limiting reagent when there is 8 grams of CH4
and 18 grams of H2O
CH4(g) + H2O(g) → CO(g) + 3 H2(g)
Limiting reagent when there is 8 grams of CH4
and 18 grams of H2O
CH4(g) + H2O(g) → CO(g) + 3 H2(g)
Additional Calculations: $400Additional Calculations: $400
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CH4(g)CH4(g)
Additional Calculations: $500Additional Calculations: $500
Answer
4.00 g of the gas was produced in a flask
containing 24.8 g of the reactant. Determine the
percent yield of the student's reaction.
4.00 g of the gas was produced in a flask
containing 24.8 g of the reactant. Determine the
percent yield of the student's reaction.
Additional Calculations: $500Additional Calculations: $500
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30.3%30.3%
Titration CalculationTitration Calculation
Proceed
Answer
3 NH4NO3 + Na3PO4 (NH4)3PO4 + 3 NaNO3
a) With 30 g of NH4NO3 and 50 g of Na3PO4, identify the limiting reagent.
b) What is the maximum amount of each product that can be formed?
3 NH4NO3 + Na3PO4 (NH4)3PO4 + 3 NaNO3
a) With 30 g of NH4NO3 and 50 g of Na3PO4, identify the limiting reagent.
b) What is the maximum amount of each product that can be formed?
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NH4NO3NH4NO3
a) Identify the limiting reagent
nNH4NO3
nNa3PO4 =
0.375 mol NH4NO3 produces 0.125 mol of (NH4)3PO4
0.305 mol of Na3PO4 produces 0.305 mol of (NH4)3PO4
NH4NO3 is the limiting reagent b) What is the maximum amount of each product that can be formed?
Since NH4NO3 is the limiting reagent, only 0.125 mol of (NH4)3PO4 and 0.375 mol of NaNO3 can be produced.