Answer:
About 16.42 grams
Explanation:
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How many litres (L) of oxygen at STP can be obtained from 110 g of potassium chlorate? [R=0.0821 L.atm/K.mol]
1 answer:
Balance the equation first:
2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)
Moles of KClO3 = 110 / 122.5 = 0.89
Following the balanced chemical equation:
We can say moles of O2 produce =
x moles of KClO3
So, O2 = (3 / 2) x 0.89
= 1.34 moles
So, Volume at STP = nRT / P
T = <span>273.15 K
P = 1 atm
So, V = (1.34 x 0.0821 x 273.15) / 1 = 30.2 L</span>
2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)
Moles of KClO3 = 110 / 122.5 = 0.89
Following the balanced chemical equation:
We can say moles of O2 produce =
So, O2 = (3 / 2) x 0.89
= 1.34 moles
So, Volume at STP = nRT / P
T = <span>273.15 K
P = 1 atm
So, V = (1.34 x 0.0821 x 273.15) / 1 = 30.2 L</span>
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With the Lucas test, tertiary alcohols react immediately producing turbidity, while secondary alcohols do so in five minutes. Primary alcohols do not react significantly with Lucas reagent at room temperature.
2. No reaction (See the attached drawing)
3. (see the attached drawing)
