Assume that magnesium would act atom-for-atom the same as copper in this experiment. How many grams of magnesium would have been
1 answer:
Answer:
0.1127 grams of Mg is used to produce one gram of silver
Explanation:
Write a balanced equation
Mg(s) +2Ag + (aq) --> (aq) + 2Ag(s)
The atomic mass of magnesium is 24.31 g/mol and the atomic mass of silver is 107.84g/mol
1 g Ag of silver is produced from [(1 mol Mg)(24.31 g/mol)] /[ (2mol Ag)(107.84g/mol)]
1 g Ag of silver is produced from 0.1127 grams of Mg
0.1127 grams of Mg is used to produce one gram of silver
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Taking into account the reaction stoichiometry, the correct answer is the third option: 15.63 moles of HgO are needed to produce 250 g of O₂.
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:
- HgO: 2 moles
- Hg: 2 moles
- O₂: 1 moles
The molar mass of the compounds is:
- HgO: 216.59 g/mole
- Hg: 200.59 g/mole
- O₂: 32 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- HgO: 2 moles× 216.59 g/mole= 433.18 grams
- Hg: 2 moles× 200.59 g/mole= 401.18 grams
- O₂: 1 mole× 32 g/mole= 32 grams
Then the following rule of three can be applied: if by reaction stoichiometry 32 grams of O₂ are produced by 2 moles of HgO, 250 grams of O₂ are produced from how many moles of HgO?
<u><em>moles of HgO= 15.625 moles≅ 15.63 moles</em></u>
Finally, the correct answer is the third option: 15.63 moles of HgO are needed to produce 250 g of O₂.
Learn more about reaction stoichiometry:
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