Answer:
[tex]m_{O_2}=120gO_2[/tex]
Explanation:
Hello!
In this case, since the decomposition of potassium chlorate is:
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
We can see a 2:3 mole ratio between potassium chlorate and oxygen (molar mass 32.0 g/mol), thus, via stoichiometry, we compute the mass of oxygen that are produced by the decomposition of 2.50 moles of this reactant:
[tex]m_{O_2}=2.50molKClO_3*\frac{3molO_2}{2molKClO_3} *\frac{32.0gO_2}{1molO_2}\\\\m_{O_2}=120gO_2[/tex]
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