Question

2. A 7.11 gram sample of Potassium Chlorate decomposed during the following reaction:

2KClO3 --> 2KCl + 3O2

How many moles of oxygen gase were formed during this process ?

2. A 7.11 gram sample of Potassium Chlorate decomposed during the following reaction:

2KClO3 --> 2KCl + 3O2

How many moles of oxygen gase were formed during this process ?

0.0870 mol

0.0293 mol

0.0278 mol

0.0665 mol

Answer 1

To find out how many moles of oxygen gas are formed from the decomposition of 7.11 grams of Potassium Chlorate, we calculate the molar mass of KClO3, convert grams of KClO3 to moles, and then use stoichiometry to determine the moles of O2 produced, resulting in 0.0870 mol of oxygen gas.

To determine how many moles of oxygen gas were formed in the decomposition of Potassium Chlorate (KClO3), we need to follow a step-by-step stoichiometric calculation:

Calculate the molar mass of KClO3: K (39.10 g/mol) + Cl (35.45 g/mol) + 3*O (3 * 16.00 g/mol) = 122.55 g/mol.

Determine the moles of KClO3 we have: 7.11 g / 122.55 g/mol = 0.0580 mol KClO3.

Use the balanced equation to find the ratio of KClO3 to O2: 2 moles of KClO3 produce 3 moles of O2. Thus, 0.0580 mol KClO3 will produce 3/2 * 0.0580 mol O2 = 0.0870 mol O2.

The calculation shows that the correct answer is 0.0870 mol of oxygen gas formed.