Directions: Write the balanced equation for each of the following situations. SHOW ALL OF YOUR WORK ON ATTACHED PAGES, OR IT WIL

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Directions: Write the balanced equation for each of the following situations. SHOW ALL OF YOUR WORK ON ATTACHED PAGES, OR IT WIL

L NOT BE ACCEPTED. 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 1. 5.00 x 1015 ng of potassium chlorate is heated to form potassium chloride and oxygen 1. Balanced Chemical Equation: Reaction Type:At the completion of reactions: Grams of potassium chlorate: Grams of potassium chloride: Grams of oxygen:

Chemistry

1 answer:

AveGali [126] · Answer 1 · 2024-03-20T06:06:21+03:00

Balanced chemical equation of potassium chlorate heated to form potassium chloride and oxygen:

$2KClO_{3(s)}+\text{heat}\rightarrow2KCl_{(s)}+3O_{2(g)}$

Reaction type: Decomposition reaction.

Now we will use the stoichiometry to calculate the masses of reactants and products after reaction completion.

We are given that the mass of KClO3 = 5.00x10^15 ng = 5000000 g

We will first calculate the number of moles of KClO3 and use the stoichiometry to calculate other masses.

number of moles (n) of KClO3:

n = m/M where m is the mass in grams and M is the molar mass in g/mol

n = (5000000 g)/(122,55g/mol)

n = 40799.67 mol of KClO3

Mass of KCl:

To find the number of moles of KCl, we will use the molar ratio between KClO3:KCl which is 2:2 according to the equation. Therefore number of moles of KCl = 40799.67 mol

m = nM

m = (40799.67 mol) x (74,5513 g/mol)

m = 3041668.707 g

Mass of Oxygen:

We will use the molar ratio between KClO3:O2 which is 2:3, therefore the number of moles of O2 = 40799.67 x (3/2) = 61199.505 mol

m = nM

m = (61199.505 mol) x (15.999 x 2)

m = 1958261.761 g

There should be NO mass of KClO3 after the reaction.

However, initially we had a mass of 5000000 g of KClO3 before the reaction. After the reaction a mass of 3041668.707 g KCl and 1958261.761 g O2 was produced whuch in total is = 4999930.468 g

The mass of KClO3 left = 5000000 g - 4999930.468 g = 69.532 g