48g of 02 produce how many grams of Al2O3
1 answer:
Taking into account the reaction stoichiometry, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
4 Al + 3 O₂ → 2 Al₂O₃
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:
- Al: 4 moles
- O₂: 3 moles
- Al₂O₃: 2 moles
The molar mass of the compounds is:
- Al: 27 g/mole
- O₂: 32 g/mole
- Al₂O₃: 102 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al: 4 moles ×27 g/mole= 108 grams
- O₂: 3 moles ×32 g/mole= 96 grams
- Al₂O₃: 2 moles ×102 g/mole= 204 grams
The following rule of three can be applied: if by reaction stoichiometry 96 grams of O₂ form 204 grams of Al₂O₃, 48 grams of O₂ form how much mass of Al₂O₃?
<u><em>mass of Al₂O₃= 102 grams</em></u>
Finally, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
Learn more about the reaction stoichiometry:
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Answer:
HCO₂
Explanation:
From the information given:
The mass of the elements are:
Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g
To determine the empirical formula;
First thing is to find the numbers of moles of each atom.
For Carbon:
For Hydrogen:
For Oxygen:
Now; we use the smallest no of moles to divide the respective moles from above.
For carbon:
For Hydrogen:
For Oxygen:
Thus, the empirical formula is HCO₂