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When 46.0 g KClO3 is heated 12.1 g O2 is formed how many grams of KCl is also formed?
Wiki User
∙ 9y ago
The equation that describes this process is as follows:
2KClO3 ---> 2KCl + 3O2
For every 2 moles of reactants consumed 3 moles of oxygen gas are produced.
3 mol O2 / 2 mol KClO3 = x mol O2 / 12.3 mol KClO3
x = 12.3 mol x 3 mol / 2 mol = 18.45 mol
Therefore, 18.5 mol (3 significant figures) of oxygen are produced by the decomposition of 12.3 mol of potassium chlorate
Wiki User
∙ 8y ago
Wiki User
∙ 8y agoThe reaction is:
2 KClO3 = 2 KCl + 3 O2
18,45 moles of oxygen are evolved.
Wiki User
∙ 8y ago2KClO3==>2KCl+3O2 is the equation. so you need 4 moles of KClO3.
Wiki User
∙ 7y agoThe reaction is:
2 KClO3 = 2 KCl + 3 O2 (three moles of oxygen)
Wiki User
∙ 8y agoThe answer is 27,98 g KCl.
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Equation: 2KClO3 + Cl2 ---> 2KCl + 3O2 + Cl2 1. Solve for the number of moles of KClO3 in 36.3 g. (.2962 molKClO3) 2. Multiply that value by (3/2), from the equation's coefficients. (.4447 molO2) Note: A BCA table could also be used. 3. Solve for the mass of .4447 molO2. 14.2 grams of oxygen would be produced.
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