Answer:
[tex]m_{O_2}=87.2gO_2[/tex]
Explanation:
Hello.
In this case, given the chemical reaction, we can compute the grams of oxygen by using the 98.2 g of water via the 2:1 mole ratio between them, the molar mass of water that is 18.02 g/mol, the molar mass of gaseous oxygen that is 32.00 g/mol and the following stoichiometric procedure relating the given information:
[tex]m_{O_2}=98.2gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molO_2}{2molH_2O}*\frac{32.00gO_2}{1molO_2} \\\\m_{O_2}=87.2gO_2[/tex]
In which the result is displayed with three significant figures because the given mass of water 98.2 g, has three significant figures too.
Best regards!