Question

1. 2KCIO, 2KCI + 30₂ If 500 g of KCIO, decomposes and produces 303 g of KCI, how many grams of O₂ are produced?​

Answer 1

To find the amount of oxygen gas produced in the decomposition of potassium chlorate (KCIO,), we can use the balanced chemical equation and the given mass of KCIO,.

Step-by-step explanation:

This question is asking us to determine the amount of oxygen gas produced in the decomposition of potassium chlorate (KCIO,). The balanced chemical equation for this reaction is:

2KCIO, → 2KCI + 30,

In order to find the amount of oxygen gas produced, we need to use the molar ratio between KCIO, and O,. From the balanced equation, we can see that 2 moles of KCIO, produce 3 moles of O,. To find the number of moles of KCIO, in 500 g, we need to use its molar mass. The molar mass of KCIO, is 122.55 g/mol, so the number of moles of KCIO, is:

500 g / 122.55 g/mol = 4.081 mol

Using the molar ratio, we can find the number of moles of O, produced:

4.081 mol KCIO, × (3 mol O,/2 mol KCIO,) = 6.1215 mol O,

To find the mass of O, produced, we can use its molar mass. The molar mass of O, is 32.00 g/mol, so the mass of O, produced is:

6.1215 mol O, × 32.00 g/mol = 195.892 g O,

Therefore, 195.892 g of O, are produced in the decomposition of 500 g of KCIO,.

Answer 2

Answer:195 g of O2 are produced when 500 g of KClO3 decompose and produce 303 g of KCl.

Explanation: The balanced chemical equation for the decomposition of 2KClO3 into 2KCl and 3O2 is:

2KClO3 → 2KCl + 3O2

According to the equation, for every 2 moles of KClO3 that decompose, 3 moles of O2 are produced.

We can use this information to set up a proportion to find the amount of O2 produced when 500 g of KClO3 decompose and produce 303 g of KCl:

2 moles KClO3 / 303 g KCl = 3 moles O2 / x g O2

where x is the amount of O2 produced.

First, we need to convert the mass of KCl to moles using its molar mass:

Molar mass KCl = 39.1 g/mol + 35.5 g/mol = 74.6 g/mol

303 g KCl / 74.6 g/mol = 4.06 moles KCl

Now we can solve for x:

2 moles KClO3 / 4.06 moles KCl = 3 moles O2 / x

Cross-multiplying and solving for x, we get:

x = (3 moles O2 * 4.06 moles KCl) / 2 moles KClO3

x = 6.09 moles O2

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

Molar mass O2 = 2(16.0 g/mol) = 32.0 g/mol

6.09 moles O2 * 32.0 g/mol = 195 g O2

Therefore, approximately 195 g of O2 are produced when 500 g of KClO3 decompose and produce 303 g of KCl.