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Gaseous hydrogen and oxygen can be prepared in the laboratory from the decomposition of gaseous water. The equation for the reaction is:

\[ 2H_2O(g) \rightarrow 2H_2(g) + O_2(g) \]

Calculate how many grams of \[ O_2(g) \] can be produced from 98.2 g of \[ H_2O(g) \].

Answer :

Answer:

[tex]m_{O_2}=87.2gO_2[/tex]

Explanation:

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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.

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