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 12.9 grams of O2 b 29.0 grams of O2 c 79.4 grams of O2 d 116.1 grams of O2
To find the amount of O2 produced when 74.0g of KClO3 decomposes, we first need to calculate the amount of moles of KClO3 in 74.0g:
Molar mass of KClO3 = 122.4 g/mol
74.0g / 122.4 g/mol = 0.604 moles of KClO3
From the balanced chemical equation for the decomposition of KClO3, we know that 2 moles of KClO3 produce 3 moles of O2.
So, for 0.604 moles of KClO3:
0.604 moles KClO3 x (3 moles O2 / 2 moles KClO3) = 0.906 moles of O2
Now, we can convert moles of O2 to grams:
Molar mass of O2 = 32 g/mol
0.906 moles O2 x 32 g/mol = 29.0 grams of O2
Therefore, the correct answer is b) 29.0 grams of O2.