1)Consider the decomposition of hydrogen peroxide into liquid water and oxygen g

Question

1)Consider the decomposition of hydrogen peroxide into liquid water and oxygen gas:

2 H202 (AQ) -> 2H20 (L) + 02 (G)

A) HOW MANY OF EACH OF THE PRODUCT MOLECULES WILL BE FORMED FROM THE REACTION OF TWO DOZEN MOLECULES OF H202? 4 DOZEN?

B) HOW MANY OF THE PRODUCT MOLECULES WILL BE FORMED FROM THE REACTION OF 2 MILLION MOLECULES OF H202? 4 MILLION?

C) HOW MANY OF EACH OF THE PRODUCT MOLECULES WILL BE FORMED FROM THE REACTION OF 2 MOLES OF H202? 4 MOLES? 15 MOLES?

D) WHAT IS THE RELATIONSHIP AMONG YOUR ANSWERS TO THE QUESTIONS ABOVE? EXPLAIN WHY THIS RELATIONSHIP EXISTS.

2) BALANCE THE EQUATION C3H80 (L) + 02 (G) --> CO2 (G) + H20 (G)

A) HOW MANY MOLES OF C3H80 WILL REACT WITH 2.500 MOLES OF OXYGEN GAS?

BALANCE THE EQUATION FE (S) + 02 (G) --> FE203 (S)

A) HOW MANY MOLES OF IRON WILL REACT WITH 8.40 MOLES OF OXYGEN GAS?

Explanation / Answer

1)   2 H2O2 (AQ) -> 2 H2O (L) + 1 O2 (G)

A) HOW MANY OF EACH OF THE PRODUCT MOLECULES WILL BE FORMED FROM THE REACTION OF TWO DOZEN MOLECULES OF H202? 4 DOZEN?    We see a 2:2+1 ratio with reactant:products. So, for two dozen molecules, we would form 2 dozen molecules of H2O + 1 dozen molecules O2.    @ 4 dozen, we have 4 dozen water and 2 dozen diatomic oxygen.

B) HOW MANY OF THE PRODUCT MOLECULES WILL BE FORMED FROM THE REACTION OF 2 MILLION MOLECULES OF H202? 4 MILLION?     Same concept as Part A. 2:2+1 ratio. @2 million H2O2, we form 2 million molecules of water and 1 million molecules of diatomic oxygen. @ 4 million H2O2, we form 4 million molecules of water and 2 million molecules of diatomic oxygen.

C) HOW MANY OF EACH OF THE PRODUCT MOLECULES WILL BE FORMED FROM THE REACTION OF 2 MOLES OF H202? 4 MOLES? 15 MOLES? This one is slightly different because we are going to have to convert moles to molecules. It starts out the same though. 2:2+1 ratio. This time, @4 moles of H2O2, we form 4 moles of water and 2 moles of diatomic oxygen. 4 moles of water * 6.02*1023 =2.41 * 1024 molecules of water and 2 moles * 6.02*1023 = 1.2 * 1024      @15 moles of H2O2, we form 15 moles of water and 7.5 moles of diatomic oxygen. 15 moles * 6.02*1023 =9.03 * 1023 and 7.5 moles * 6.02*1023 = 4.52 * 1024 molecules.

D) WHAT IS THE RELATIONSHIP AMONG YOUR ANSWERS TO THE QUESTIONS ABOVE? EXPLAIN WHY THIS RELATIONSHIP EXISTS.     The coefficients in a chemical equation can represent either Molecular ratio or Molar ratio. They cannot be used at the same time in an equation though, without conversion using Avagrado's number.

2) BALANCE THE EQUATION C3H80 (L) + 02 (G) --> CO2 (G) + H20 (G)
2 C3H8O + 9 O2 --> 6 CO2 + 8 H2O

A) HOW MANY MOLES OF C3H80 WILL REACT WITH 2.500 MOLES OF OXYGEN GAS? Molar ratio of reactants is 2:9. So for 2.5 moles of oyxgen, we need to divide by oxygen's molar ratio and multiply by propanol's molar ratio. 2.5 / 9 * 2 = 0.5 moles of propanol.

BALANCE THE EQUATION FE (S) + 02 (G) --> FE203 (S)
2 Fe (s) + 3/2 O2 (g) --> Fe2O3 (s)       Step 1
4 Fe (s) + 3 O2 (g) --> 2 Fe2O3 (s)        Final Step

A) HOW MANY MOLES OF IRON WILL REACT WITH 8.40 MOLES OF OXYGEN GAS?
Molar ratio is 4:3     number of moles of oxygen divided by oxygen molar ratio multiplied by iron molar ratio.      8.40 / 3 * 4 = 11.2 moles of Iron.

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