2NaBrO3 reacts to 2NaBr + 3O2
How many grams of NaBrO3 are required to produce 19 grams of O3?
How many liters of 02 are produced from 4 moles of NaBrO3?
How many atoms of NaBr are produced when 1 mole of O2 is produced?
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To answer these questions, you need to use the balanced chemical equation and convert between moles and grams using molar mass. Here's how you can solve each question step by step:
1. How many grams of NaBrO3 are required to produce 19 grams of O3?
First, you need to find the molar mass of O3 (ozone). Oxygen (O) has a molar mass of 16.00 g/mol, and since O3 has three oxygen atoms, the molar mass of O3 is 16.00 g/mol x 3 = 48.00 g/mol.
Using the balanced chemical equation 2NaBrO3 -> 2NaBr + 3O2, you can see that for every 3 moles of O2 produced, you need 2 moles of NaBrO3.
Now, calculate the number of moles of O3 present in 19 grams. Using the molar mass of O3 (48.00 g/mol), you can calculate the number of moles using the formula: Moles = Mass / Molar mass.
Moles of O3 = 19 g / 48.00 g/mol = 0.3958 mol.
According to the balanced equation, for every 3 moles of O2 produced, 2 moles of NaBrO3 are needed. Therefore, the number of moles of NaBrO3 required to produce 0.3958 mol of O3 is:
Moles of NaBrO3 = (2/3) x 0.3958 mol = 0.2638 mol.
Now, to find the grams of NaBrO3 required, you can multiply the number of moles by the molar mass of NaBrO3. The molar mass of NaBrO3 is found by summing the molar masses of each element: 22.99 g/mol (Na) + 79.90 g/mol (Br) + 48.00 g/mol (O) x 3 = 150.89 g/mol.
Grams of NaBrO3 = 0.2638 mol x 150.89 g/mol = 39.82 grams.
Therefore, you would need approximately 39.82 grams of NaBrO3 to produce 19 grams of O3.
2. How many liters of O2 are produced from 4 moles of NaBrO3?
Using the balanced chemical equation, you can see that for every 2 moles of NaBrO3, 3 moles of O2 are produced.
Since you have 4 moles of NaBrO3, you can set up a ratio to find the moles of O2 produced:
(4 moles NaBrO3) x (3 moles O2 / 2 moles NaBrO3) = 6 moles O2.
To convert moles to liters, you need to use the Ideal Gas Law, which states that at standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.
So, 6 moles of O2 would occupy:
6 moles O2 x 22.4 L/mol = 134.4 liters of O2.
Therefore, 4 moles of NaBrO3 would produce 134.4 liters of O2.
3. How many atoms of NaBr are produced when 1 mole of O2 is produced?
Using the balanced chemical equation, you can see that for every 2 moles of NaBrO3, 2 moles of NaBr are produced, and for every 3 moles of O2, 2 moles of NaBr are produced.
Therefore, the ratio of moles of O2 to moles of NaBr is 3:2.
Since you have 1 mole of O2, you can set up a proportion to find the moles of NaBr:
(1 mole O2) x (2 moles NaBr / 3 moles O2) = 0.67 moles NaBr.
To find the number of atoms of NaBr, you need to use Avogadro's number, which states that 1 mole of any substance contains 6.022 x 10^23 particles.
So, 0.67 moles of NaBr would contain:
0.67 moles NaBr x 6.022 x 10^23 atoms/mole = 4.03 x 10^23 atoms of NaBr.
Therefore, when 1 mole of O2 is produced, approximately 4.03 x 10^23 atoms of NaBr are also produced.