Taking into account the reaction stoichiometry and the definition of molar mass, the quantity of HgO used is 0.01385 moles and 0.006925 moles of O₂ are formed when 3.00 grams of HgO was used.
In first place, the balanced reaction is:
2 HgO → 2 Hg + O₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The molar mass of the compounds is:
You know that 3.00 grams of HgO was used. Then, the quantity in moles of HgO used, can be calculated using the molar mass of HgO as follow:
[tex]3 gramsx\frac{1 mole}{216.59 grams}[/tex]= 0.01385 moles
Finally, the quantity of HgO used is 0.01385 moles.
To determine the number of moles of oxygen gas formed, the following rule of three can be applied: if by reaction stoichiometry 2 moles of HgO form 1 mole of O₂, 0.01385 moles of HgO form how many moles of O₂?
[tex]amount of moles of O_{2} =\frac{0.01385 moles of HgOx1 mole of O_{2} }{2 moles of HgO}[/tex]
amount of moles of O₂= 0.006925 moles
Then, 0.006925 moles of O₂ are formed when 3.00 grams of HgO was used.
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