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.
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
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