Question

Potassium chlorate (used in fireworks, flares and safety matches) forms oxygen and potassium chloride

when heated.
KClO3(s) → KCl(s) + O2(g)
How many grams of oxygen are formed when 26.4 g of potassium chlorate is heated?
A. 4.60 g
B. 6.86 g
C. 99.1 g
D. 10.3 g
E. 223 g

Answer 1

B. 6.86 g.

Step-by-step explanation:

To solve this problem, we need to use stoichiometry. The balanced chemical equation tells us that 1 mole of KClO3 decomposes to form 1 mole of O2. We can use the molar mass of KClO3 to convert the given mass of KClO3 to moles, and then use the mole ratio from the balanced equation to find the moles of O2 produced. Finally, we can use the molar mass of O2 to convert moles of O2 to grams.

The molar mass of KClO3 is calculated as follows:

K = 1(39.10 g/mol) = 39.10 g/mol

Cl = 1(35.45 g/mol) = 35.45 g/mol

O = 3(16.00 g/mol) = 48.00 g/mol

Molar mass of KClO3 = 39.10 + 35.45 + 48.00 = 122.55 g/mol

Now we can use the given mass of KClO3 to find the moles of KClO3:

moles of KClO3 = 26.4 g / 122.55 g/mol = 0.215 mol

From the balanced chemical equation, we know that 1 mole of KClO3 produces 1 mole of O2. Therefore, the number of moles of O2 produced is also 0.215 mol.

Finally, we can convert the moles of O2 to grams using the molar mass of O2:

mass of O2 = 0.215 mol x 32.00 g/mol = 6.86 g

Therefore, the answer is B. 6.86 g.