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2.76 g of silver carbonate on being strongly heated yields a residue weighing
- a)2.16 g
- b)2.48 g
- c)2.32 g
- d)2.64 g
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
2.76 g of silver carbonate on being strongly heated yields a residue w...
2Ag2CO3(s) ----> 4Ag(s) + 2CO2(g) + O2(g)
So, on strogly heating silver carbonate will produce silver.
No of moles of silver carbonate = 2.76 g/ 275.7 = 0.0100
No of moles of Ag = x g/ 108 =
275.7 g of Ag2CO3 produces 108 x 2 g of Ag.
2.76 g will produce = 108 x 2 x 2.76 / 275.7 = 2.16 g of silver.
Most Upvoted Answer
2.76 g of silver carbonate on being strongly heated yields a residue w...
Given:
Mass of silver carbonate = 2.76 g
Residue mass = x g
To find:
The mass of residue obtained on strongly heating 2.76 g of silver carbonate.
Solution:
The chemical equation for the decomposition of silver carbonate is given as follows:
Ag2CO3 (s) → 2Ag (s) + CO2 (g) + 1/2 O2 (g)
From the above equation, we can see that silver carbonate decomposes into silver metal, carbon dioxide gas, and oxygen gas.
To calculate the mass of residue obtained on strongly heating 2.76 g of silver carbonate, we need to find the mass of silver metal produced.
We can find the mass of silver metal produced by using the molar mass of silver carbonate and the stoichiometry of the reaction.
The molar mass of silver carbonate is:
Ag2CO3 = (2 x 107.87 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol) = 275.77 g/mol
The stoichiometry of the reaction shows that one mole of silver carbonate produces two moles of silver.
So, the number of moles of silver produced can be calculated as:
Number of moles of Ag = (2.76 g Ag2CO3) / (275.77 g/mol Ag2CO3) x (2 mol Ag / 1 mol Ag2CO3) = 0.02 mol Ag
The mass of silver produced can be calculated as:
Mass of Ag = 0.02 mol Ag x 107.87 g/mol Ag = 2.16 g Ag
Therefore, the mass of residue obtained on strongly heating 2.76 g of silver carbonate is 2.16 g.
Hence, the correct option is A.
Mass of silver carbonate = 2.76 g
Residue mass = x g
To find:
The mass of residue obtained on strongly heating 2.76 g of silver carbonate.
Solution:
The chemical equation for the decomposition of silver carbonate is given as follows:
Ag2CO3 (s) → 2Ag (s) + CO2 (g) + 1/2 O2 (g)
From the above equation, we can see that silver carbonate decomposes into silver metal, carbon dioxide gas, and oxygen gas.
To calculate the mass of residue obtained on strongly heating 2.76 g of silver carbonate, we need to find the mass of silver metal produced.
We can find the mass of silver metal produced by using the molar mass of silver carbonate and the stoichiometry of the reaction.
The molar mass of silver carbonate is:
Ag2CO3 = (2 x 107.87 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol) = 275.77 g/mol
The stoichiometry of the reaction shows that one mole of silver carbonate produces two moles of silver.
So, the number of moles of silver produced can be calculated as:
Number of moles of Ag = (2.76 g Ag2CO3) / (275.77 g/mol Ag2CO3) x (2 mol Ag / 1 mol Ag2CO3) = 0.02 mol Ag
The mass of silver produced can be calculated as:
Mass of Ag = 0.02 mol Ag x 107.87 g/mol Ag = 2.16 g Ag
Therefore, the mass of residue obtained on strongly heating 2.76 g of silver carbonate is 2.16 g.
Hence, the correct option is A.
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2.76 g of silver carbonate on being strongly heated yields a residue weighinga)2.16 gb)2.48 gc)2.32 gd)2.64 gCorrect answer is option 'A'. Can you explain this answer?
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