Equation Stoichiometry 3.notebook
November 19, 2013
Equation Stoichiometry Problems 3
If exactly 7.50 g of propane gas, C3H8, reacts with 6.00 g of oxygen gas, 1.
write the BME and include the energy term:
a)
C3H8(l) + 5O2(g) à 3CO2(g) + 4H2O(g) + 2219.2 kJ
b)
calculate the mass of water produced:
7.50 g C3H8 x (1 mol C3H8/44.0 g C3H8) x (4 mol H2O/1 mol C3H8) x (18.0 g H2O/1 mol H2O) = 12.3 g H2O
6.00 g O2 x (1 mol O2/32.0 g O2) x (4 mol H2O/5 mol O2) x (18.0 g H2O/1 mol H2O) = 2.70 g H2O
Therefore O2 is limiting reactant and 2.70 g H2O is produced
c)
calculate the energy released in this reaction:
6.00 g O2 x (1 mol O2/32.0 g O2) X 2219.2 kJ/5 mol O2) = ­333 kJ
d)
calculate the mass of excess reactant that remains:
6.00 g O2 x (1 mol O2/32.0 g O2) x (1 mol C3H8/5 mol O2) x (44.0 g C3H8/1 mol C3H8) = 1.65 g C3H8 reacts
Therefore 7.50 g start – 1.65 g reacts = 5.85 g C3H8 left unreacted
Nov 17­5:44 PM
2.
In the Haber Process, nitrogen gas reacts with hydrogen gas to produce ammonia gas. If 2.00 kilograms of nitrogen gas reacts with 1.50 kilograms of hydrogen gas
a)
write the BME including the energy term:
3H2(g) + N2(g) à 2NH3(g) + 91.8 kJ
b)
determine the limiting reactant and mass of ammonia formed:
2.00 X 103 g N2 x (1 mol N2/28.0 g N2) x (2 mol NH3/1 mol N2) x (17.0 g NH3/1 mol NH3) = 2430 g NH3
1.50 X 103 g H2 x (1 mol H2/2.0 g H2) x (2 mol NH3/3 mol H2) x (17.0 g NH3/1 mol NH3) = 8500 g NH3
Therefore N2 is limiting reactant and 2430 g NH3 produced
c)
determine the mass of excess reactant remaining after the reaction stops:
2.00 X 103 g N2 x (1 mol N2/28.0 g N2) x (3 mol H2/1 mol N2) x (2.0 g H2/1 mol H2) = 429 g H2 reacts
Therefore, 1500 g H2 start – 429 g H2 reacts = 1070 g H2 left unreacted
d)
if 994 g of ammonia is actually collected, calculate the percent yield:
994/2439 X 100% = 40.9%
e)
calculate the energy involved in this reaction and indicate whether the energy was absorbed or released:
2.00 X 103 g N2 x (1 mol N2/28.0 g N2) x (91.8 kJ/1 mol N2) = 6.56 X 103 kJ
This energy is released because ∆H <0, therefore exothermic
Nov 17­5:44 PM
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Equation Stoichiometry 3.notebook
3.
November 19, 2013
When carbon reacts with hydrogen gas, ethane gas is produced, having the chemical formula C2H6.
a)
write a BME and include the energy term:
2C(s) + 3H2(g) à C2H6(g) + 84.0 kJ
if 540.0 g of carbon reacts with 125.0 g of hydrogen gas,
b)
determine the limiting reactant and calculate the mass of ethane gas produced:
540.0 g C x (1 mol C/12.0 g C) x (1 mol C2H6/2 mol C) x (30.0 g C2H6/1 mol C2H6) = 675.0 g C2H6
125.0 g H2 x (1 mol H2/2.0 g H2) x (1 mol C2H6/3 mol H2) x (30.0 g C2H6/1 mol C2H6) = 625.0 g C2H6
Therefore H2 is limiting reactant and 625.0 g C2H6 produced
c)
calculate the mass of excess reactant leftover:
125.0 g H2 x (1 mol H2/2.0 g H2) x (2 mol C/3 mol H2) x (12.0 g C/1 mol c) = 500.0 g C reacts
Therefore, 540.0 g C start – 500.0 g C reacts = 40.0 g C left unreacted
d)
if 595.0 g of ethane gas is collected, calculate the percent yield:
595.0/625.0 X 100% = 95.2%
e)
calculate the energy involved in this reaction and indicate whether the energy was absorbed or released:
125.0 g H2 x (1 mol H2/2.0 g H2) x (84.0 kJ/3 mol H2) = 1750 kJ
Nov 17­5:45 PM
4.
An aqueous solution containing 4.50 g of lead(II)nitrate is mixed with an aqueous solution containing 3.15 g of potassium sulfate.
a)
write a BME:
Pb(NO3)2(aq) + K2SO4(aq) à PbSO4(s) + 2 KNO3(aq)
b)
calculate the mass of precipitate formed:
4.50 g Pb(NO3)2 x (1 mol Pb(NO3)2/331.2 g Pb(NO3)2) x (1 mol PbSO4/1 mol Pb(NO3)2) x (303.3 g PbSO4/1 mol PbSO4) = 4.12 g PbSO4
3.15 g K2SO4 x (1 mol K2SO4/174.3 g K2SO4) x (1 mol PbSO4/1 mol K2SO4) x (303.3 g PbSO4/1 mol PbSO4) = 5.48 g PbSO4
Therefore Pb(NO3)2 is limiting reactant and 4.12 g PbSO4 is produced
c)
calculate the mass of excess reactant that remains:
4.50 g Pb(NO3)2 x (1 mol Pb(NO3)2/331.2 g Pb(NO3)2) x (1 mol K2SO4/1 mol Pb(NO3)2) x (174.3 g K2SO4/1 mol K2SO4) = 2.36 g K2SO4 reacts
Therefore, 3.15 g K2SO4 start – 2.36 g K2SO4 reacts = 0.79 g K2SO4 left unreacted
Nov 17­5:46 PM
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Equation Stoichiometry 3.notebook
5.
November 19, 2013
A reaction between aqueous solutions of calcium nitrate and lithium sulfide resulted in the production of 12.80 g of calcium sulfide. If this mass represented a 75.0% yield, calculate the mass of calcium nitrate that was required to produce 100% yield. Assume the lithium sulfide was in excess.
Ca(NO3)2(aq) + Li2S(aq) à CaS(s) + 2LiNO3(aq
7.50/100 = 12.80/X, X = 17.07 g CaS
17.07 g CaS x (1 mol CaS/72.2 g CaS) x (1 mol Ca(NO3)2/1 mol CaS) x 9164.1 g Ca(NO3)2/1 mol Ca(NO3)2) = 38.80 g Ca(NO3)2
Nov 17­5:47 PM
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