Asked by UnName
Identify the following chemical equations by type.
1. If an electric discharge produces 500 cm3 of ozone (O3), how many cm3 of oxygen (O2) are required?
3O2(g) ---> 2O3(g)
2. When 2.75 dm3 of O2 react with an excess of glucose (C6H12O2), according to the reaction below, what volume of carbon dioxide will be produced?
6O2(g) + C6H12O6(s) ---> 6H2O(g) + 6CO2(g)
3. If an excess of nitrogen gas reacts with 25.0 L of hydrogen gas, according to the reaction below, how many L of ammonia will be produced?
N2(g) + 3H2(g) ---> 2NH3(g)
4. How many cm3 of oxygen would be required to react comp
1. If an electric discharge produces 500 cm3 of ozone (O3), how many cm3 of oxygen (O2) are required?
3O2(g) ---> 2O3(g)
2. When 2.75 dm3 of O2 react with an excess of glucose (C6H12O2), according to the reaction below, what volume of carbon dioxide will be produced?
6O2(g) + C6H12O6(s) ---> 6H2O(g) + 6CO2(g)
3. If an excess of nitrogen gas reacts with 25.0 L of hydrogen gas, according to the reaction below, how many L of ammonia will be produced?
N2(g) + 3H2(g) ---> 2NH3(g)
4. How many cm3 of oxygen would be required to react comp
Answers
Answered by
DrBob222
1. If an electric discharge produces 500 cm3 of ozone (O3), how many cm3 of oxygen (O2) are required?
3O2(g) ---> 2O3(g)
<b> This is a synthesis. Since these are gases we can use cc as if they were mols.
500 cc O3 x (3 mols O2/2 mols O3) = 5-- x 3/2 = ?
</b>
3O2(g) ---> 2O3(g)
<b> This is a synthesis. Since these are gases we can use cc as if they were mols.
500 cc O3 x (3 mols O2/2 mols O3) = 5-- x 3/2 = ?
</b>
Answered by
DrBob222
#2 and #3 are worked the same way as #1.
#2 is combustion.
#3 is synthesis
#4 doesn't make sense to me.
#2 is combustion.
#3 is synthesis
#4 doesn't make sense to me.
Answered by
UnName
Answers;
1.
(500 cm^3 O3)*(3 moles O2/2 moles O3) = 500*(3/2) cm^3 O2 = 750 cm^3 O2
2)
(2.75 dm3 O2) x (6 dm3 CO2 / 6 dm3 O2) = 2.75 dm3 CO2
3)
(25.0 L H2) x (2 L NH3 / 3 L H2) = 16.7 L NH3
4)
(250.0 cm3 H2) x (2 cm3 H2 / 1 cm3 O2) = 500.0 cm3 O2
1.
(500 cm^3 O3)*(3 moles O2/2 moles O3) = 500*(3/2) cm^3 O2 = 750 cm^3 O2
2)
(2.75 dm3 O2) x (6 dm3 CO2 / 6 dm3 O2) = 2.75 dm3 CO2
3)
(25.0 L H2) x (2 L NH3 / 3 L H2) = 16.7 L NH3
4)
(250.0 cm3 H2) x (2 cm3 H2 / 1 cm3 O2) = 500.0 cm3 O2
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