Question

Directions: Write the balanced equation for each of the following situations. . In addition, list the reaction type. YOU MUST TELL THE AMOUNTS OF EVERY SUBSTANCE THAT REMAINS IN THE CONTAINER AT THE END OF THE REACTION. ASSUME THAT ALL REACTIONS GO TO COMPLETION. If only STOICHIOMETRY, tell how much of the excess reactant is used!!!! Reaction Type a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction 5. 2.11 mg of hydrogen peroxide decomposes in the presence of manganese dioxide to form water and oxygen 5. Balanced Chemical Equation Reaction Type:At the completion of reactions: Grams of hydrogen peroxide: Grams of water: Grams of oxygen gas:

Answer 1

First, let's write the balanced chemical reaction:

2 H2O2 ---MnO2--> 2 H2O + O2

This reaction is a b. Decomposition Reaction of hydrogen peroxide.

Step 1. Let's transform 2.11 mg of H2O2 into moles using the following equation mole = mass / molar mass

Step 2. We are going to use the equation proportion to calculate the quantity in moles of each compound.

Step 3. We are going to transform moles into grams of each compound using the following equation: mass = mole x molar mass

Step 1.

We need to transform 2.11 mg to g. We just need to divide by 1000,

So: 2.11 x 10^-3 g of H2O

mole = 2.11 x 10^-3/34.02

mole = 6.20 x 10^-5 moles

Step 2.

2 moles of H2O2 ---- 2 moles of H2O

6.20 x 10^-5 moles ---- x moles of H2O

x = 6.20 x 10^-5 moles of H2O

2 moles of H2O2 ---- 1 moles of O2

6.20 x 10^-5 moles ---- x moles of O2

x = 3.10 x 10^-5 moles of O2

Step 3.

molar mass of O2 : 32 g/mol

molar mass of H2O: 18 g/mol

mass of H2O = 6.20 x 10^-5 x 18 = 1.12 x 10^-3 g

mass of O2 = 3.10 x 10^-5 x 32 = 9.92 x 10^-4 g