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Stoichiometry Word Problems 2 SOLUTIONS. 1. Cellular respiration occurs in animal cells, a reaction that is essentially the combustion of a sugar.
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Student Name: _______________________________________________ Pd. ____ Date: __________
Balanced Equation: C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O
6 12 6 (^61262)
6 12 6 2
6 12 6 2
gCH O molCH O
gCH O molO
molCH O LO
L O molO =
Balanced Equations: 4 C 3 H 5 (ONO 2 ) 3 10 H 2 O + 12 CO 2 + 6 N 2 + O 2
molgas
Lgas molnitro
molgas gnitro
molnitro kgnitro
kg nitro gnitro =
b. How much energy is produced by the explosion?
molnitro
MJenergy gnitro
molnitro kgnitro
kg nitro gnitro
Balanced Equation: 2 NaCl + 2 H 2 O 2 NaOH + H 2 + Cl 2
gH O molHO
gHO molNaCl
molHO gNaCl
g NaCl molNaCl 2 2
Since the amount of H 2 O required is less than the amount provided (4562 g), it is in excess, so NaCl is the limiting reagent.
b. How much base is produced?
Use the limiting reactant (NaCl):
g NaOH molNaOH
gNaOH molNaCl
molNaOH gNaCl
g NaCl molNaCl
c. How much of the excess reactant is left over?
From part (a), 721.54 g of water were consumed. Subtract this from the given (4562 – 721.54 = 3840 g H 2 O are left over.
2 NH 4 Cl + 2 CaO CaCl 2 + Ca(OH) 2 + 2 NH 3
If 75.6 g of ammonium chloride is allowed to react with 52.8 g of quicklime: a. How many grams of the excess reactant are left over?
g CaO molCaO
gCaO molNHCl
molCaO gNHCl
g NHCl molNHCl
4 4
Since more quicklime is needed than is provided, quicklime is the limiting reactant. Therefore:
gNH Cl molNHCl
gNHCl molCaO
molNHCl gCaO
g CaO molCaO 4 4
75.6 – 40.4 = 35.2 grams of NH 4 Cl are left over.
b. How many grams of ammonia are formed?
Use the limiting reactant:
3 3
gNH molNH
gNHCl molCaO
molNH gCaO
g CaO molCaO =