How many grams of oxygen, O2 are produced if 74.0g of potassium chlorate, KClO3, decomposes?
The molar mass of O2 = 32 g/mol and the molar mass of KClO3 = 122.4 g/mol).
If you are unsure how to convert from grams to grams, please see the example problem linked HERE.
a
116.1 grams of O2
b
29.0 grams of O2
c
79.4 grams of O2
d
12.9 grams of O2
To find the amount of oxygen produced when 74.0g of potassium chlorate decomposes, we first need to determine the molar ratio between KClO3 and O2.
From the balanced chemical equation for the decomposition of potassium chlorate:
2KClO3 -> 2KCl + 3O2
This shows that 2 moles of KClO3 produce 3 moles of O2.
Now, we can calculate the number of moles of KClO3 in 74.0g using the molar mass:
74.0g / 122.4 g/mol = 0.604 moles of KClO3
With the molar ratio, we can find the number of moles of O2 produced:
0.604 moles KClO3 * (3 moles O2 / 2 moles KClO3) = 0.906 moles O2
Finally, we convert moles of O2 to grams of O2 using the molar mass:
0.906 moles O2 * 32 g/mol = 28.99g
Therefore, the correct answer is b) 29.0 grams of O2.